General Redox Reaction Concepts
Redox reactions, or reduction-oxidation reactions, involve the transfer of electrons between substances. These reactions are central to numerous chemical processes, including electrochemistry, organic transformations, and biological energy production. Mastery of redox concepts is essential for JEE Advanced preparation, as these reactions form the basis of many complex problem-solving scenarios.
1. Oxidation and Reduction: Definitions and Oxidation States
· Oxidation: Defined as the loss of electrons or an increase in oxidation state. When a substance undergoes oxidation, its oxidation number increases.
- Example: In the reaction \( \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \), iron is oxidized as it loses electrons.
· Reduction: Defined as the gain of electrons or a decrease in oxidation state. A substance is reduced when its oxidation number decreases.
- Example: In \( \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \), chlorine is reduced as it gains electrons.
· Oxidizing and Reducing Agents:
- Oxidizing Agent: Accepts electrons and is reduced in the reaction.
- Reducing Agent: Donates electrons and is oxidized in the reaction.
Assigning Oxidation States
The rules for assigning oxidation numbers are critical for complex redox reactions, especially in organic and transition metal compounds:
- Pure elements have an oxidation state of zero.
- For monatomic ions, the oxidation state equals the ion charge.
- In compounds, oxygen generally has an oxidation state of -2, and hydrogen is typically +1.
- The sum of oxidation states in a neutral compound is zero, and in a polyatomic ion, it equals the ion’s charge.
Example: Determine the oxidation states in \( \text{HNO}_3 \):
- \( \text{H} = +1\), \( \text{O} = -2\) (for each oxygen), and to balance, \( \text{N} = +5\).
2. Balancing Redox Reactions
JEE Advanced questions frequently require balancing redox reactions, especially in acidic or basic solutions. Two main methods are used:
1. Oxidation State Method:
- Identify changes in oxidation states for elements involved in the reaction.
- Equalize the total increase and decrease in oxidation numbers by adjusting coefficients.
- Balance other elements, and finally, balance hydrogen and oxygen by adding \( \text{H}^+ \) (for acidic medium) or \( \text{OH}^- \) (for basic medium).
2. Half-Reaction Method:
- Separate the redox reaction into oxidation and reduction half-reactions.
- Balance each half-reaction for mass and charge.
- Add the half-reactions, ensuring that the electrons lost in oxidation equal the electrons gained in reduction.
Example: Balance \( \text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Mn}^{2+} + \text{CO}_2 \) in acidic solution:
- Oxidation half: \( \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2e^- \)
- Reduction half: \( \text{MnO}_4^- + 8 \text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \)
- Multiply and combine to balance the reaction.
3. Types of Redox Reactions and Advanced Applications
1. Combination Reactions: Two or more elements/compounds combine to form a product. E.g., \( \text{H}_2 + \text{Cl}_2 \rightarrow 2 \text{HCl} \).
2. Decomposition Reactions: A single compound breaks down. E.g., \( 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \).
3. Displacement Reactions: An element in a compound is replaced by another element. E.g., \( \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu} \).
Applications:
- Metallurgy: Reduction of metal ores, like iron extraction from hematite.
- Batteries: Redox reactions in galvanic cells generate electric power, e.g., the lead-acid battery.
- Biological Systems: Cellular respiration involves electron transfer reactions that release energy.
4. Electrochemical Cells and Nernst Equation
Electrochemical cells are essential for converting chemical energy into electrical energy:
- Galvanic Cells: Where spontaneous redox reactions occur to produce electric current.
- Anode: Oxidation occurs here (negative).
- Cathode: Reduction occurs here (positive).
- The cell potential (E°) is calculated from standard reduction potentials.
- Electrolytic Cells: Where an external voltage drives a non-spontaneous reaction.
The Nernst Equation adjusts cell potential for non-standard conditions:
\[
E = E^\circ - \frac{RT}{nF} \ln Q
\]
Where:
- \(E^\circ\) is the standard cell potential,
- \(R\) is the gas constant,
- \(T\) is the temperature in Kelvin,
- \(n\) is the number of moles of electrons,
- \(F\) is the Faraday constant, and
- \(Q\) is the reaction quotient.
Example: Calculating cell potential for a Zn-Cu cell under non-standard conditions.
5. Practice Problems and Example Calculations
1. Oxidation State Calculations: Determine the oxidation state of elements in complex compounds like \( \text{K}_2\text{Cr}_2\text{O}_7 \) and \( \text{KMnO}_4 \).
2. Redox Reactions in Acid/Base Media: Practice balancing reactions in acidic and basic environments to reinforce method proficiency.
3. Cell Potential Calculations: Use the Nernst Equation to calculate potentials for given conditions, especially for concentration cells and electrolysis setups.
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Q 1. What is ' \(A\) ' in the following reaction \(2 \mathrm{Fe}^{3+}(a q)+\mathrm{Sn}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{2+}(a q)+\mathrm{A}\); |
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(d) \(\mathrm{Sn}\); |
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(c) \(\mathrm{Sn}^{2+}(a q)\); |
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(b) \(\mathrm{Sn}^{4+}(a q)\); |
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(a) \(\mathrm{Sn}^{3+}(a q)\); |
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Q 2. Given \(\mathrm{E}^{\Theta}\) (i) \(\mathrm{Mg}^{2+} / \mathrm{Mg}(\mathrm{s}), \mathrm{E}^{\Theta}=-2.36\) (ii) \(\mathrm{Ag}^{+} / \mathrm{Ag}(\mathrm{s}), \mathrm{E}^{\Theta}=0.80\) (iii) \(\mathrm{Al}^{3+} / \mathrm{Al}(\mathrm{s}), \mathrm{E}^{\Theta}=-1.66\) (iv) \(\mathrm{Cu}^{2+} / \mathrm{Cu}(\mathrm{s}), \mathrm{E}^{\Theta}=0.52\) Out of the above given elements which is the strongest oxidising agent and which is the weakest oxidising agent?; |
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(d) (ii) is the strongest whereas (iii) is the weakest oxidising agent; |
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(c) (i) is the strongest whereas (ii) is the weakest oxidising agent; |
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(b) (ii) is the strongest whereas (i) is the weakest oxidising agent; |
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(a) (iv) is the strong whereas (ii) is the weakest oxidising agent; |
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Q 3. Stronger is oxidising agent, more is; |
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(d) standard oxidation potential of that species; |
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(c) the tendency to lose electrons by that species; |
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(b) the tendency to get it self oxidised; |
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(a) standard reduction potential of that species; |
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Q 4. Electrode potential data are given below: \[ \begin{align*} &\mathrm{Fe}_{(\mathrm{aq})}^{+3} + \mathrm{e}^{-} \longrightarrow \mathrm{Fe}_{(\mathrm{aq})}^{+2} ; & \mathrm{E}^{\circ} &= +0.77 \mathrm{~V} \\ &\mathrm{Al}_{(\mathrm{aq})}^{3+} + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}_{(\mathrm{s})} ; & \mathrm{E}^{\circ} &= -1.66 \mathrm{~V} \\ &\mathrm{Br}_{2(\mathrm{aq})} + 2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}_{(\mathrm{aq})}^{-} ; & \mathrm{E}^{\circ} &= +1.08 \mathrm{~V} \\ \end{align*} \] Based on the data, the reducing power of \(\mathrm{Fe}^{2+}, \mathrm{Al},\) and \(\mathrm{Br}^{-}\) will increase in the order: ; |
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(d) \(\mathrm{Al}<\mathrm{Fe}^{2+}<\mathrm{Br}^{-}\); |
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(c) \(\mathrm{Al}<\mathrm{Br}^{-}<\mathrm{Fe}^{2+}\); |
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(b) \(\mathrm{Fe}^{2+}<\mathrm{Al}<\mathrm{Br}^{-}\); |
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(a) \(\mathrm{Br}^{-}<\mathrm{Fe}^{2+}<\mathrm{Al}\); |
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Q 5. Oxidation numbers of \(\mathrm{P}\) in \(\mathrm{PO}_{4}^{3-}\), of \(\mathrm{S}\) in \(\mathrm{SO}_{4}^{2-}\) and that of \(\mathrm{Cr}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) are respectively; |
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(a) \(+3,+6\) and +5; |
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(b) \(+5,+3\) and +6; |
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(c) \(-3,+6\) and +6; |
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(d) \(+5,+6\) and +6; |
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Q 6. Which of the following statement(s) is/are correct ? (i) All alkali metals and some alkaline earth metals \((\mathrm{Ca}\), \(\mathrm{Sr}\) and \(\mathrm{Ba}\) ) displace hydrogen from cold water. (ii) Magnesium and iron react with steam as well as acids to produce hydrogen gas. (iii) Cadmium and tin do not react with steam but displace hydrogen from acids.; |
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(a) (i) and (ii); |
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(b) (ii) only; |
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(c) (i) and (iii); |
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(d) (i), (ii) and (iii); |
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Q 7. Assertion : \(\mathrm{HClO}_{4}\) is a stronger acid than \(\mathrm{HClO}_{3}\) Reason : Oxidation state of \(\mathrm{Cl}\) in \(\mathrm{HClO}_{4}\) is \(+\mathrm{VII}\) and in \(\mathrm{HClO}_{3}\) \(+\mathrm{V}\).; |
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(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
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(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
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(c) Assertion is correct, reason is incorrect; |
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(d) Assertion is incorrect, reason is correct.; |
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Q 8. \(\mathrm{E}^{\ominus}\) Values of some redox couples are given below. On the basis of these values, choose the correct option: \[ \begin{align*} &\mathrm{E}^{\ominus} \text{ values :} \\ &\mathrm{Br}_{2} / \mathrm{Br}^{-} = +1.90 \\ &\mathrm{Ag}^{+} / \mathrm{Ag}(\mathrm{s}) = +0.80 \\ &\mathrm{Cu}^{2+} / \mathrm{Cu}(\mathrm{s}) = +0.34 \\ &\mathrm{I}_{2}(\mathrm{s}) / \mathrm{I} = +0.54 \\ \end{align*} \] ; |
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(a) Cu will reduce \(\mathrm{Br}^{-}\); |
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(b) Cuwill reduce \(\mathrm{Ag}\); |
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(c) Cu will reduce \(\mathrm{I}^{-}\); |
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(d) \(\mathrm{Cu}\) will reduce \(\mathrm{Br}_{2}\); |
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Q 9. The volume (in \(\mathrm{mL}\) ) of \(0.1 \mathrm{NNaOH}\) required to neutralise \(10 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) phosphinic acid is; |
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(D)20 mL; |
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(C)12 mL; |
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(B)10 mL; |
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(A)6 mL; |
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Q 10. When \(10 \mathrm{~mL}\) of an aqueous solution of \(\mathrm{Fe}^{2+}\) ions was titrated in the presence of dil \(\mathrm{H}_{2} \mathrm{SO}_{4}\) using diphenylamine indicator, \(15 \mathrm{~mL}\) of \(0.02 \mathrm{M}\) solution of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) was required to get the end point. The molarity of the solution containing \(\mathrm{Fe}^{2+}\) ions is \(\mathrm{x} \times 10^{-2} \mathrm{M}\). The value of \(\mathrm{x}\) is . (Nearest integer); |
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(D)36; |
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(C)18; |
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(B)10; |
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(A)9; |
Oxidation Reaction
Oxidation reactions are foundational to redox chemistry, where a substance loses electrons, resulting in an increase in its oxidation state. Understanding oxidation in depth is crucial for JEE Advanced, as it is integral to topics like electrochemistry, metallurgy, organic transformations, and biochemical energy cycles. In this note, we’ll dive deeper into oxidation principles, complex balancing techniques, applications, and examples.
1. Understanding Oxidation and Oxidation States
· Definition: Oxidation is defined as the loss of electrons by an atom, molecule, or ion, resulting in an increase in its oxidation number.
- Example: In the reaction \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \), zinc undergoes oxidation as it loses electrons.
· Oxidation State Rules: For advanced problem-solving, remember key rules:
- Pure elements have an oxidation state of 0.
- For ions, the oxidation state equals the charge.
- In compounds, common oxidation states include \( -2 \) for oxygen (except in peroxides) and \( +1 \) for hydrogen (except in hydrides).
- Example: In \( \text{Cr}_2\text{O}_7^{2-} \), assign each oxygen as -2, yielding a total of -14 for seven oxygens, so chromium must be +6 each to balance the -2 charge.
· Oxidation vs. Reduction: Oxidation always occurs with a paired reduction reaction (in a redox process), where the oxidizing agent gains the electrons that are lost.
2. Types of Oxidation Reactions
· Combination Reactions: When elements or compounds combine, one is oxidized.
- Example: \( 2 \text{Fe} + \text{O}_2 \rightarrow 2 \text{FeO} \), where iron is oxidized to form iron oxide.
· Decomposition Reactions: A compound breaks down, and one component undergoes oxidation.
- Example: \( 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \), where chlorine is reduced, while oxygen is liberated.
· Single Displacement Reactions: A more reactive metal displaces another from its compound.
- Example: \( \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu} \), where zinc displaces copper by oxidizing itself.
· Combustion Reactions: These high-energy reactions involve a hydrocarbon reacting with oxygen, releasing energy and yielding CO\(_2\) and H\(_2\)O.
- Example: \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \); carbon is oxidized, oxygen is reduced.
3. Oxidizing Agents and Advanced Applications
· Common Strong Oxidizing Agents:
- Potassium Permanganate (\(\text{KMnO}_4\)): A powerful oxidizer in acidic medium, converting \(\text{Mn}^{7+}\) to \(\text{Mn}^{2+}\).
- Potassium Dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7\)): Widely used in labs, it oxidizes various organic and inorganic species, reducing \(\text{Cr}^{6+}\) to \(\text{Cr}^{3+}\).
- Chlorine (\(\text{Cl}_2\)): Used in water treatment and organic transformations, oxidizes metals and nonmetals alike.
· Applications:
- Industrial Processes: Oxidation is essential in metal extraction (e.g., smelting), bleaching, and synthesis of chemicals like sulfuric acid.
- Biological Systems: Cellular respiration involves the oxidation of glucose, providing energy for cellular functions.
- Batteries and Fuel Cells: In batteries, oxidation reactions generate electric current, as in lithium-ion and lead-acid cells.
4. Balancing Oxidation Reactions: Advanced Techniques
Mastering balancing, especially in acidic and basic media, is essential for JEE Advanced. The Half-Reaction Method provides a structured approach to balancing complex reactions.
· Balancing in Acidic Medium:
- Example: Balance \( \text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} \) in acidic solution.
- Steps:
- Oxidation Half-Reaction: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \)
- Reduction Half-Reaction: \( \text{MnO}_4^- + 8 \text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \)
- Multiply the oxidation half by 5 to equalize electrons.
- Combine and simplify to get:
\[\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\]
· Balancing in Basic Medium:
- Example: \( \text{Cr(OH)}_3 + \text{ClO}^- \rightarrow \text{CrO}_4^{2-} + \text{Cl}^- \).
- Steps:
- Separate into half-reactions and balance elements other than hydrogen and oxygen.
- Use \(\text{OH}^-\) and \(\text{H}_2\text{O}\) to balance H and O atoms.
- Add and simplify to achieve a balanced reaction in basic conditions.
5. Advanced Applications of Oxidation in Electrochemistry
· Electrochemical Cells:
- Galvanic Cells: Spontaneous oxidation occurs at the anode, generating electric current. For example, in a Zn-Cu cell:
- Anode reaction (oxidation): \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
- Cathode reaction (reduction): \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)
- Electrons flow from Zn to Cu, generating electricity.
· Electrolytic Cells: Non-spontaneous oxidation reactions occur when an external voltage is applied.
- Example: In electrolysis of water, oxygen is generated at the anode (oxidation), and hydrogen at the cathode (reduction).
· Nernst Equation:
· The Nernst equation allows calculation of cell potentials for redox reactions under non-standard conditions: \[E = E^\circ - \frac{RT}{nF} \ln Q\]
- Example: Calculate the potential for a Zn-Cu cell with non-standard concentrations.
1. Oxidation State Calculation:
- Determine the oxidation states of each element in \( \text{K}_2\text{Cr}_2\text{O}_7 \) and balance any reaction involving its reduction to \( \text{Cr}^{3+} \).
2. Complex Redox Balancing:
- Balance the reaction \( \text{MnO}_4^- + \text{SO}_3^{2-} \rightarrow \text{Mn}^{2+} + \text{SO}_4^{2-} \) in acidic solution using the half-reaction method.
3. Electrochemical Cell Calculations:
- Calculate the cell potential for a galvanic cell composed of \( \text{Fe}^{3+}/\text{Fe}^{2+} \) and \( \text{Ag}^{+}/\text{Ag} \) half-cells, using the Nernst equation for non-standard concentrations.
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Q 1. \(\mathrm{Co}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Co}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) The above reaction is; |
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(d) None of these; |
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(c) redox reaction; |
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(b) reduction reaction; |
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(a) oxidation reaction; |
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Q 2. One mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\) loses 10 moles of electrons to form a new compound, y Assuming that all nitrogen appear in the new compound, what is the oxidation state of nitrogen in y (There is no change in the oxidation state of hydrogen ); |
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(d) +5; |
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(c) +3; |
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(b) -3; |
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(a) -1; |
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Q 3. What is the oxidation number of elements in the free or in the uncombined state ?; |
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(d) -1; |
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(c) +2; |
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(b) 0; |
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(a) +1; |
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Q 4. In which of the following compounds oxygen has highest oxidation state and in which it has lowest oxidation state? \(\mathrm{OF}_{2}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{KO}_{2}, \mathrm{O}_{2} \mathrm{~F}_{2}\); |
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(d) Highest \(=\mathrm{KO}_{2}\), lowest \(=\mathrm{H}_{2} \mathrm{O}_{2}\); |
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(c) Highest \(=\mathrm{OF}_{2}\), lowest \(=\mathrm{KO}_{2}\); |
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(b) Highest \(=\mathrm{OF}_{2}\), lowest \(=\mathrm{K}_{2} \mathrm{O}_{2}\); |
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(a) Highest \(=\mathrm{KO}_{2}\), lowest \(=\mathrm{H}_{2} \mathrm{O}_{2}\); |
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Q 5. Oxidation number of \(\mathrm{N}\) in \(\mathrm{HNO}_{3}\) is; |
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(d) +5; |
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(c) -5; |
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(b) +3.5; |
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(a) -3.5; |
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Q 6. In which of the following reactions, there is no change in valency?; |
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(d) \(3 \mathrm{BaO}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{BaO}_{2}\).; |
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(c) \(\mathrm{BaO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{BaSO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2}\); |
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(b) \(\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{~S} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{~S}\); |
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(a) \(4 \mathrm{KClO}_{3} \longrightarrow 3 \mathrm{KClO}_{4}+\mathrm{KCl}\); |
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Q 7. The oxidation number of chromium in potassium dichromate is; |
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(a) +6; |
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(b) -5; |
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(c) -2; |
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(d) +2; |
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Q 8. The oxidation number of sulphur in \(\mathrm{S}_{8}, \mathrm{~S}_{2} \mathrm{~F}_{2}, \mathrm{H}_{2} \mathrm{~S}\) respectively, are; |
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(a) \(0,+1\) and -2; |
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(b) \(+2,+1\) and -2; |
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(c) \(0,+1\) and +2; |
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(d) \(-2,+1\) and -2; |
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Q 9. In which of the following compounds, iron has lowest oxidation state?; |
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(a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\); |
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(b) \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\); |
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(c) \(\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\); |
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(d) \(\mathrm{Fe}(\mathrm{CO})_{5}\); |
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Q 10. The oxidation state of osmium \((\mathrm{Os})\) in \(\mathrm{OsO}_{4}\) is; |
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(a) +7; |
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(b) +6; |
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(c) +4; |
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(d) +8; |
Reducing Agents in Redox Reactions
A reducing agent (or reductant) in a redox reaction is the species that donates electrons to another species, causing it to reduce while itself getting oxidized. The reducing agent undergoes an increase in oxidation state by losing electrons, thereby enabling the reduction of the other reactant.
Characteristics of a Reducing Agent
1. Electron Donor: A reducing agent donates electrons to another substance. By losing electrons, it is oxidized.
- Example: In the reaction \( \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu} \), zinc acts as the reducing agent as it loses electrons to copper ions, causing copper to reduce.
2. Oxidation State Change: During a redox reaction, the oxidation state of the reducing agent increases.
- Example: In the reaction \( \text{Mg} + \text{O}_2 \rightarrow \text{MgO} \), magnesium’s oxidation state changes from 0 to +2 as it is oxidized, making it the reducing agent.
3. Relation to Electrochemical Series: The strength of a reducing agent can be understood by its position in the electrochemical series. Strong reducing agents, like alkali metals and alkali earth metals, are highly reactive and tend to lose electrons easily.
Types of Reducing Agents and Their Applications
1. Metals: Many metals, including sodium, potassium, magnesium, and zinc, serve as reducing agents due to their tendency to lose electrons readily.
- Example: In metallurgy, carbon or carbon monoxide is often used to reduce metal oxides to pure metals, like in the reduction of iron oxide in a blast furnace.
2. Non-metals and Compounds: Compounds like hydrogen, carbon, and carbon monoxide are also effective reducing agents.
- Example: In organic chemistry, hydrogen gas is used to reduce alkenes to alkanes in catalytic hydrogenation processes.
3. Biological Reducing Agents: In biological systems, NADH and FADH\(_2\) act as reducing agents in cellular respiration, providing electrons to the electron transport chain.
- Example: In cellular respiration, NADH donates electrons to the mitochondrial electron transport chain, leading to ATP production.
Advanced Examples and Balancing with Reducing Agents
1. Balancing Redox Reactions Using Half-Reactions:
- Example Problem: Balance \( \text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} \).
- Oxidation Half-Reaction (Iron): \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \)
- Reduction Half-Reaction (Permanganate): \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)
- Balanced Reaction: \( \text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \)
2. Special Role of Reducing Agents in Displacement Reactions: In single displacement reactions, a more reactive metal (stronger reducing agent) displaces a less reactive metal from its compound.
- Example: \( \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu} \)
- Here, zinc is oxidized, acting as a reducing agent, while copper is reduced.
3. Electrochemical Cells:
- In galvanic cells, reducing agents are essential at the anode, where oxidation occurs. For instance, in a zinc-copper cell, zinc serves as the reducing agent, releasing electrons and causing the cell’s current flow.
Strength and Selection of Reducing Agents
1. Electronegativity and Reducing Power: Elements with low electronegativity (like alkali and alkaline earth metals) are usually strong reducing agents as they readily donate electrons.
2. Comparative Reducing Strength: In the electrochemical series, metals higher up (like lithium) are stronger reducing agents than those lower down (like copper), making them more effective for specific reduction processes.
- Example: Lithium and sodium are strong reducing agents used in specialized industrial processes, including lithium batteries.
3. Reactivity with Water and Air: Strong reducing agents like sodium react even with water and oxygen in the air.
- Example: Sodium metal reacts vigorously with water to produce hydrogen gas, showcasing its strong reducing ability.
Applications in Industry and Biochemistry
1. Metallurgy: Reducing agents such as carbon and hydrogen are crucial in metal extraction.
- Example: Carbon reduces iron oxides in blast furnaces to produce pure iron.
2. Organic Chemistry: Reducing agents like lithium aluminum hydride (LiAlH\(_4\)) are used for reducing carbonyl compounds.
- Example: LiAlH\(_4\) reduces ketones and aldehydes to alcohols in organic synthesis.
3. Biochemistry: In metabolic processes, reducing agents such as NADH and FADH\(_2\) donate electrons, crucial for ATP generation through oxidation in the electron transport chain.
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Q 1. In reaction, \(4 \mathrm{Na}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{O}\), sodium behaves as; |
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(a) oxidising agent; |
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(b) reducing agent; |
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(c) Both (a) and (b); |
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(d) None of these; |
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Q 2. Nitric oxide acts as a reducing agent in the reaction; |
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(a) \(4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}\); |
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(b) \(2 \mathrm{NO}+3 \mathrm{I}_{2}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NO}_{3}^{-}+61^{-}+8 \mathrm{H}^{+}\); |
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(c) \(2 \mathrm{NO}+\mathrm{H}_{2} \mathrm{SO}_{3} \rightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{SO}_{4}\); |
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(d) \(2 \mathrm{NO}+\mathrm{H}_{2} \mathrm{~S} \rightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{S}+\mathrm{H}_{2} \mathrm{O}\); |
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Q 3. A negative \(\mathrm{E}^{\odot}\) means that redox couple is a \(\qquad\)A than the \(\mathrm{H}^{+} / \mathrm{H}_{2}\) couple A positive \(\mathrm{E}^{\ominus}\) means that the redox couple is a \(\qquad\) B than \(\mathrm{H}^{+} / \mathrm{H}_{2}\) couple; |
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(a) \(\mathrm{A}=\) stronger reducing agent \(\mathrm{B}=\) weaker reducing agent; |
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(b) \(\mathrm{A}=\) stronger oxidising agent \(\mathrm{B}=\) weaker oxidising agent; |
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(c) \(\mathrm{A}=\) weaker oxidising agent \(\mathrm{B}=\) stronger oxidising agent; |
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(d) Both (a) and (c); |
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Q 4. Which of the following statement(s) is/are correct? (i) A negative value of \(\mathrm{E}^{-}\)means that the redox couple is a weaker reducing agent than the \(\mathrm{H}^{+} / \mathrm{H}_{2}\) couple. (ii) A positive \(\mathrm{E}^{-}\)means that the redox couple is weaker reducing agent than the \(\mathrm{H}^{+} / \mathrm{H}_{2}\). Which of the following code is incorrect regarding above statements?; |
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(a) Only (i); |
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(b) only (ii); |
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(c) Both (i) and (ii); |
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(d) Neither (i) nor (ii); |
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Q 5. Assertion : In the reaction \(2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})\) sodium is oxidised Reason : Sodium acts as an oxidising agent in given reaction.; |
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(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
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(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
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(c) Assertion is correct, reason is incorrect; |
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(d) Assertion is incorrect, reason is correct.; |
Redox Reaction
A redox reaction is one in which oxidation and reduction processes occur simultaneously, involving the transfer of electrons between reactants. Redox reactions play a crucial role in various chemical processes, including energy production, corrosion, and biological systems, making them essential for JEE Advanced chemistry.
1. Fundamental Concepts of Redox Reactions
· Oxidation: The process in which an atom, ion, or molecule loses electrons, resulting in an increase in oxidation number.
- Example: \( \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \), where iron undergoes oxidation.
· Reduction: The process in which an atom, ion, or molecule gains electrons, resulting in a decrease in oxidation number.
- Example: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), where copper ion undergoes reduction.
· Oxidizing and Reducing Agents: In a redox reaction, the oxidizing agent is the substance that gains electrons and is reduced, while the reducing agent donates electrons and is oxidized.
- Example: In the reaction \( \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu} \), zinc acts as the reducing agent, and copper(II) sulfate acts as the oxidizing agent.
· Oxidation Numbers: To determine redox changes, the concept of oxidation numbers is essential. These are assigned based on specific rules to track the transfer of electrons.
2. Balancing Redox Reactions
Balancing redox reactions is essential to ensure the conservation of mass and charge. Two primary methods are used:
· Oxidation Number Method: This involves tracking changes in oxidation states to balance the reaction.
- Steps:
- Identify the oxidation states of all elements in the reaction.
- Determine which elements undergo oxidation and reduction.
- Balance changes in oxidation numbers by adding appropriate coefficients.
· Half-Reaction Method: Commonly used for reactions in aqueous solutions, this method separates the oxidation and reduction reactions.
- Steps:
· Write the oxidation and reduction half-reactions.
· Balance each half-reaction for all atoms except hydrogen and oxygen.
· Add \( \text{H}_2\text{O} \) to balance oxygen and \( \text{H}^+ \) ions to balance hydrogen.
· Balance the charges by adding electrons.
· Multiply each half-reaction by an appropriate factor to equalize electron numbers, then combine.
· Example: Consider balancing the reaction between permanganate and iron in acidic solution:
- \( \text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} \)
- Oxidation Half-Reaction: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \)
- Reduction Half-Reaction: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)
- Combine to yield: \( \text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \)
3. Types of Redox Reactions
· Combination Reactions: Elements or compounds combine to form a product.
- Example: \( \text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl} \), where hydrogen is oxidized and chlorine is reduced.
· Decomposition Reactions: A compound decomposes into simpler compounds or elements.
- Example: \( 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \), where \( \text{ClO}_3^- \) is reduced to \( \text{Cl}^- \).
· Displacement Reactions: A more reactive element displaces another in a compound.
- Example: \( \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu} \).
· Disproportionation Reactions: An element undergoes both oxidation and reduction in the same reaction.
- Example: \( 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \), where oxygen is both oxidized and reduced.
4. Applications of Redox Reactions
· Electrochemistry: Redox reactions drive electrochemical cells, including galvanic and electrolytic cells. In galvanic cells, spontaneous redox reactions generate electrical energy.
- Example: In a Daniell cell, zinc is oxidized at the anode, releasing electrons, while copper ions are reduced at the cathode.
· Biological Processes: Redox reactions are vital in cellular respiration and photosynthesis, where electron transfer drives ATP synthesis.
· Industrial Applications: Redox reactions are used in metallurgy (e.g., extraction of iron) and in the production of chemicals like chlorine and hydrogen through electrolysis.
5. Key Points for JEE Advanced
1. Calculating Oxidation Numbers: Practicing the assignment of oxidation numbers to complex compounds and ions is crucial.
2. Balancing Complex Reactions: Mastery of both the oxidation number and half-reaction methods is necessary for tackling intricate redox questions.
3. Electrochemical Series: Familiarity with the electrochemical series helps predict reaction spontaneity and the strength of oxidizing and reducing agents.
4. Advanced Problem Solving: JEE Advanced often includes problems requiring deep understanding of redox behavior under varying conditions, such as different pH or concentration.
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Q 1. In the reaction given below, identify the species undergoing redox reaction \[ 2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NaH}(s) \]; |
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(a) \(\mathrm{Na}\) is reduced and hydrogen is oxidised; |
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(b) \(\mathrm{Na}\) is oxidised and hydrogen is reduced; |
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(c) \(\mathrm{Na}\) undergoes oxidation and hydrogen undergoes reduction; |
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(d) Both (b) and (c); |
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Q 2. Balance the following redox reactions by ion - electron method \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \quad\) (in acidic solution) by selecting any of the choices correctly; |
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(A) One moles of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) oxidises 3 moles of \(\mathrm{SO}_{2}\); |
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(B) One mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produces 3 moles of \(\mathrm{SO}_{4}^{2-}\); |
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(C) One moles of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) oxidises 3 moles of \(\mathrm{SO}_{2}\)e mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produces 3 moles of \(\mathrm{SO}_{4}^{2-}\); |
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(D) One mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) ion oxidises 2 moles of \(\mathrm{SO}_{2}\); |
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Q 3. Balance the following redox reactions by ion - electron method \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \quad\) (in acidic solution) by selecting any of the choices correctly; |
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(A) One moles of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) oxidises 3 moles of \(\mathrm{SO}_{2}\); |
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(B) One mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produces 3 moles of \(\mathrm{SO}_{4}^{2-}\); |
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(C) One moles of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) oxidises 3 moles of \(\mathrm{SO}_{2}\)e mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produces 3 moles of \(\mathrm{SO}_{4}^{2-}\); |
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(D) One mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) ion oxidises 2 moles of \(\mathrm{SO}_{2}\); |
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Q 4. Which of the following statement(s) is/are correct for the given reaction? \(2 \mathrm{HgCl}_{2}(a q)+\mathrm{SnCl}_{2}(a q) \rightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+\mathrm{SnCl}_{4}(a q)\) (i) Mercuric chloride is reduced to \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (ii) Stannous chloride is oxidised to stannic chloride (iii) \(\mathrm{HgCl}_{2}\) is oxidised to \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (iv) It is an example of redox reaction; |
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(a) (i), (ii) and (iv); |
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(b) (i) and (ii); |
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(c) (iii) and (iv); |
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(d) (iii) only; |
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Q 5. In the redox reaction, \(x \mathrm{KMnO}_{4}+\mathrm{NH}_{3} \longrightarrow y \mathrm{KNO}_{3}+\mathrm{MnO}_{2}+\mathrm{KOH}+\mathrm{H}_{2} \mathrm{O}\); |
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(a) \(x=4, y=6\); |
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(b) \(x=3, y=8\); |
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(c) \(x=8, y=6\); |
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(d) \(x=8, y=3\); |
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Q 6. Given: \(\mathrm{X} \mathrm{Na}_{2} \mathrm{HAsO}_{3}+\mathrm{YNaBrO}_{3}+\mathrm{ZHCl} \rightarrow \mathrm{NaBr}\) \(+\mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{NaCl}\) The values of \(\mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) in the above redox reaction are respectively; |
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(a) \(2,1,2\); |
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(b) \(2,1,3\); |
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(c) \(3,1,6\); |
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(d) \(3,1,4\); |
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Q 7. The values of \(x\) and \(y\) in the following redox reaction \(\mathrm{x} \mathrm{Cl}_{2}+6 \mathrm{OH}^{-} \longrightarrow \mathrm{ClO}_{3}^{-}+\mathrm{yCl}^{-}+3 \mathrm{H}_{2} \mathrm{O}\) are; |
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(a) \(x=5, y=3\); |
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(b) \(x=2, y=4\); |
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(c) \(\mathrm{x}=3, \mathrm{y}=5\); |
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(d) \(x=4, y=2\); |
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Q 8. Assertion : The reaction : \(\mathrm{CaCO}_{3}(\mathrm{~s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is an example of decomposition reaction Reason : Above reaction is not a redox reaction.; |
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(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
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(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
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(c) Assertion is correct, reason is incorrect; |
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(d) Assertion is incorrect, reason is correct.; |
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Q 9. Assertion : In a reaction \(\mathrm{Zn}(s)+\mathrm{CuSO}_{4}(a q) \rightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{Cu}(s)\) \(\mathrm{Zn}\) is a reductant but itself get oxidized. Reason : In a redox reaction, oxidant is reduced by accepting electrons and reductant is oxidized by losing electrons.; |
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(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
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(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
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(c) Assertion is correct, reason is incorrect; |
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(d) Assertion is incorrect, reason is correct.; |
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Q 10. For the redox reaction [2018] \(\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) The correct coefficients of the reactants for the balanced equation are & \(\mathrm{MnO}_{4}^{-}\) & \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) & \(\mathrm{H}^{+}\) \\; |
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(a) & 16 & 5 & 2 \\; |
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(b) & 2 & 5 & 16 \\; |
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(c) & 5 & 16 & 2 \\; |
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(d) & 2 & 16 & 5; |
DISPLACEMENT REACTIONS
1. Basic Concept
- Definition: Chemical reaction where a more reactive element displaces a less reactive element from its compound
- Type: Belongs to redox reactions (involves electron transfer)
- General equation: A + BC → AC + B (where A is more reactive than B)
2. Types of Displacement Reactions
a) Metal displacement
- Metal + Metal salt → New metal + New metal salt
- Examples:
* Zn + CuSO₄ → ZnSO₄ + Cu
* Fe + CuSO₄ → FeSO₄ + Cu
- Reactivity series plays crucial role: K > Na > Ca > Mg > Al > Zn > Fe > Pb > Cu > Hg > Ag > Au
JEE Advanced Example (2015):
Question: When iron nails are added to copper sulphate solution:
(A) Temperature of solution decreases
(B) Blue color fades
(C) pH of solution increases
(D) Brown deposit forms
Solution:
- Reaction: Fe + CuSO₄ → FeSO₄ + Cu
- Blue color fades as Cu²⁺ ions are consumed
- pH increases as H⁺ concentration decreases
- Brown Cu metal deposits
Answer: Options B, C, and D are correct
b) Non-metal displacement
- Non-metal + Non-metal compound → New non-metal compound + New non-metal
- Examples:
* Cl₂ + 2KBr → 2KCl + Br₂
* Br₂ + 2KI → 2KBr + I₂
- Reactivity order: F₂ > Cl₂ > Br₂ > I₂
3. Advanced Concepts
a) Metal displacement from acids
JEE Advanced Example (2019):
Question: In the reaction of copper metal with concentrated HNO₃, one of the products is NO₂. The other main product formed is:
Solution:
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Answer: Cu(NO₃)₂
b) Electrochemical aspects
- Based on Standard Electrode Potential (E°)
- Reaction occurs if E°(cell) is positive
- ΔG = -nFE°
JEE Advanced Example (2018):
Question: Calculate cell potential for Zn strip in CuSO₄ solution
Given: Zn²⁺/Zn = -0.76V and Cu²⁺/Cu = +0.34V
Solution:
E°cell = E°reduction - E°oxidation = 0.34 - (-0.76) = 1.10V
4. Important Factors Affecting Displacement
a) Temperature effects
- Higher temperature generally increases reaction rate
- May affect equilibrium position
- Can change reaction mechanism
b) Concentration effects
- Higher concentration increases reaction rate
- Follows mass action law
- Important for equilibrium calculations
c) Surface area
- Larger surface area increases reaction rate
- Particularly important in heterogeneous reactions
- Powder form reacts faster than bulk metal
5. Advanced Applications
a) Industrial processes
- Metal extraction
- Purification of metals
- Manufacturing of chemicals
b) Analytical chemistry
JEE Advanced Example (2016):
Question: Calculate copper deposited when 2 amperes current passes through CuSO₄ for 30 minutes.
Solution:
- Q = It = 2A × 1800s = 3600 coulombs
- n(e⁻) = 3600/96500 = 0.037 moles
- Cu²⁺ + 2e⁻ → Cu
- n(Cu) = 0.037/2 = 0.0186 moles
- Mass = 0.0186 × 63.5 = 1.18g
6. Thermodynamic Considerations
a) Gibbs free energy
- ΔG = ΔH - TΔS
- Reaction is spontaneous if ΔG < 0
- Related to electrode potential: ΔG = -nFE°
b) Enthalpy changes
- Usually exothermic for metal displacement
- Heat of reaction calculations using:
* Hess's law
* Born-Haber cycle
* Standard enthalpies of formation
7. Kinetic Aspects
a) Rate law
- Rate = k[A]ᵃ[BC]ᵇ
- Order of reaction importance
- Temperature dependence follows Arrhenius equation
b) Mechanism
- Often involves electron transfer
- May include intermediate steps
- Rate determining step crucial
8. Special Cases and Exceptions
a) Apparent exceptions to reactivity series
- Passivation (Al, Cr)
- Complex formation
- Oxide layer formation
JEE Advanced Example (2017):
Question: Arrange in increasing order of reducing power:
(i) Metal M that forms MX₂ with halogen
(ii) Metal N that displaces M from MX₂
(iii) X⁻
Solution:
- N can displace M → N is stronger reducing agent
- M reduces X₂ → M is stronger than X⁻
Answer: X⁻ < M < N
9. Practice Focus Areas
a) Numerical problems on:
- Stoichiometry
- Equilibrium calculations
- Electrochemical calculations
- Thermodynamic parameters
b) Conceptual questions on:
- Mechanism prediction
- Product prediction
- Condition optimization
- Multiple reactions
Key Points for JEE Advanced:
1. Common themes in questions:
- Metal displacement series
- Electrochemical calculations
- Reaction conditions
- Product prediction
2. Problem-solving approach:
- Write balanced equations
- Use standard reduction potentials
- Consider concentration effects
- Check for side reactions
Remember:
- Thorough understanding of electrode potentials is crucial
- Know reactivity series thoroughly
- Practice numerical problems extensively
- Focus on exceptions and special cases
- Understand both thermodynamic and kinetic aspects
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Q 1. Why the displacement reactions of chlorine, bromine and iodine using fluorine are not generally carried out in aqueous solution?; |
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(a) chlorine, bromine and iodine reacts with water and displace oxygen of water; |
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(b) Fluorine being very reactive attacks water and displaces oxygen of water; |
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(c) Fluorine does not react with chlorine, bromine and iodine in aqueous media; |
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(d) None of these; |
DISPROPORTIONATION REACTIONS
1. Basic Concept
- Definition: A redox reaction where same species undergoes both oxidation and reduction
- Key characteristic: Single element exists in different oxidation states
- Example: 2H₂O₂ → 2H₂O + O₂
* H₂O₂: O has -1 oxidation state
* H₂O: O has -2 oxidation state
* O₂: O has 0 oxidation state
2. Types of Disproportionation
a) In Aqueous Medium
Examples:
- Cl₂ + H₂O → HOCl + HCl
* Cl₂(0) → Cl(+1) in HOCl and Cl(-1) in HCl
JEE Advanced Example (2018):
Question: In the reaction of chlorine with cold dilute NaOH, the products formed are:
Solution:
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
- Cl₂(0) → Cl(-1) in NaCl
- Cl₂(0) → Cl(+1) in NaClO
b) In Basic Medium
Examples:
- 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O
JEE Advanced Example (2016):
Question: In the reaction of I₂ with NaOH, identify the products and oxidation states.
Solution:
3I₂ + 6OH⁻ → 5I⁻ + IO₃⁻ + 3H₂O
- I₂(0) → I(-1) in I⁻
- I₂(0) → I(+5) in IO₃⁻
3. Important Examples and Their Analysis
a) Hydrogen Peroxide
2H₂O₂ → 2H₂O + O₂
Analysis:
- O in H₂O₂: -1
- O in H₂O: -2
- O in O₂: 0
b) Phosphorous
4P + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
Analysis:
- P(0) → P(-3) in PH₃
- P(0) → P(+1) in NaH₂PO₂
4. Thermodynamic Aspects
a) Gibbs Free Energy
- ΔG must be negative for spontaneous reaction
- Affected by:
* Temperature
* Concentration
* pH of medium
b) Equilibrium Constants
JEE Advanced Example (2019):
Question: Calculate equilibrium constant for:
2H₂O₂ ⇌ 2H₂O + O₂
Given: ΔG° = -98.6 kJ/mol
Solution:
- ΔG° = -RT ln K
- K = e^(98600/8.314×298)
- K = 1.7 × 10¹⁷
5. Kinetic Considerations
a) Rate Laws
- Generally complex due to multiple steps
- Often first order with respect to disproportionating species
- May involve catalysts
b) Mechanism
- Usually involves radical intermediates
- Step-wise electron transfer
- pH dependent pathways
6. Factors Affecting Disproportionation
a) pH Effect
- Different products in acidic/basic medium
- Example: ClO⁻ disproportionation
* Acidic: 2HClO → 2H⁺ + 2Cl⁻ + O₂
* Basic: 3ClO⁻ → 2Cl⁻ + ClO₃⁻
b) Temperature Effect
- Higher temperature generally favors disproportionation
- Affects equilibrium position
- May change reaction pathway
7. Applications in Chemical Analysis
a) Qualitative Analysis
JEE Advanced Example (2017):
Question: Identify the products when KMnO₄ reacts in different pH.
Solution:
- Acidic: MnO₄⁻ → Mn²⁺
- Neutral: MnO₄⁻ → MnO₂
- Basic: MnO₄⁻ → MnO₄²⁻
b) Quantitative Analysis
- Iodometry
- Permanganometry
- Dichromate titrations
8. Important Disproportionation Reactions for JEE
a) Halogens
- Cl₂, Br₂, I₂ in alkaline medium
- Interhalogen compounds
- Hypohalites
b) Transition Metals
- Cu⁺ → Cu² + Cu⁰
- Mn³⁺ → Mn²⁺ + MnO₂
9. Practice Problems Focus
a) Numerical Type
- Equilibrium constant calculations
- pH effect calculations
- Electrode potential problems
b) Conceptual Questions
JEE Advanced Example (2020):
Question: Predict products of Cu⁺ disproportionation:
2Cu⁺ → Cu²⁺ + Cu
Solution:
- Cu⁺ is unstable
- Simultaneously oxidized to Cu²⁺ and reduced to Cu
- ΔG° is negative, hence spontaneous
10. Common Mistakes to Avoid
a) Balancing Equations
- Consider both atoms and charges
- Account for solvent participation
- Check oxidation states
b) Identifying True Disproportionation
- Must involve same element
- Must have both oxidation and reduction
- Check all oxidation states
Important Points for JEE Advanced:
1. Know standard reactions thoroughly
2. Practice oxidation state calculations
3. Understand pH effect on products
4. Master electron-half equations
5. Focus on application-based questions
Tips for Problem Solving:
1. Always check oxidation states first
2. Write balanced half-reactions
3. Consider reaction conditions
4. Look for catalysts/inhibitors
5. Analyze thermodynamic feasibility
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Q 1. Which of the following elements does not show disproportionation tendency?; |
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(a) \(\mathrm{Cl}\); |
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(b) \(\mathrm{Br}\); |
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(c) \(\mathrm{F}\); |
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(d) I; |
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Q 2. An example of a disproportionation reaction is; |
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(a) \(2 \mathrm{MnO}_{4}^{-}+10 \mathrm{I}^{-}+16 \mathrm{H}^{-} \rightarrow 2 \mathrm{Mn}^{2-}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O}\); |
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(b) \(2 \mathrm{NaBr}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{NaCl}+\mathrm{Br}_{2}\); |
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(c) \(2 \mathrm{KMnO}_{4} \rightarrow \mathrm{K}_{2} \mathrm{MnO}_{4}+\mathrm{MnO}_{2}+\mathrm{O}_{2}\); |
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(d) \(2 \mathrm{CuBr} \rightarrow \mathrm{CuBr}_{2}+\mathrm{Cu}\); |
DECOMPOSITION REACTIONS
1. Basic Concept
- Definition: Breaking down of a complex compound into simpler compounds or elements
- General equation: AB → A + B
- Types of decomposition: Thermal, Electrolytic, Photochemical
- Can be redox or non-redox reactions
2. Types of Decomposition Reactions
a) Thermal Decomposition
- Heat energy breaks chemical bonds
Examples:
* CaCO₃ → CaO + CO₂ (Non-redox)
* 2KClO₃ → 2KCl + 3O₂ (Redox)
JEE Advanced Example (2019):
Question: Predict the products when NH₄NO₃ decomposes at different temperatures:
Solution:
Below 300°C: NH₄NO₃ → N₂O + 2H₂O
Above 300°C: NH₄NO₃ → N₂ + 2H₂O + ½O₂
b) Electrolytic Decomposition
- Electrical energy breaks bonds
Examples:
* 2H₂O → 2H₂ + O₂
* 2NaCl → 2Na + Cl₂
JEE Advanced Example (2018):
Question: Calculate the mass of copper deposited when 0.5 ampere current passes through CuSO₄ solution for 1 hour.
Solution:
- Time = 3600s
- Q = It = 0.5 × 3600 = 1800C
- No. of moles of e⁻ = 1800/96500
- Mass of Cu = (1800/96500) × (63.5/2) = 0.592g
c) Photochemical Decomposition
- Light energy causes decomposition
Examples:
* 2AgBr → 2Ag + Br₂
* 2AgCl → 2Ag + Cl₂
3. Important Decomposition Reactions for JEE
a) Metal Carbonates
MCO₃ → MO + CO₂
Stability order:
Be < Mg < Ca < Sr < Ba
b) Metal Nitrates
2M(NO₃)₂ → 2MO + 4NO₂ + O₂
Exception: AgNO₃ → 2Ag + 2NO₂ + O₂
JEE Advanced Example (2017):
Question: Arrange the following in order of thermal stability:
AgNO₃, Cu(NO₃)₂, Zn(NO₃)₂
Solution: Zn(NO₃)₂ < Cu(NO₃)₂ < AgNO₃
4. Thermodynamic Aspects
a) Energy Considerations
- Most decomposition reactions are endothermic
- ΔH is positive
- ΔG must be negative for spontaneous reaction
b) Temperature Effect
- Higher temperature favors decomposition
- Follows Le Chatelier's principle
- Affects reaction rate exponentially
5. Kinetic Aspects
a) Rate Law
- Usually first order
- Rate = k[A]
- Half-life independent of initial concentration
b) Activation Energy
- High activation energy barrier
- Catalysts often needed
- Temperature dependent
6. Mechanism and Intermediates
a) Step-wise Decomposition
Example: Ammonium dichromate
(NH₄)₂Cr₂O₇ → Cr₂O₃ + N₂ + 4H₂O
JEE Advanced Example (2016):
Question: Write mechanism for thermal decomposition of H₂O₂
Solution:
Step 1: H₂O₂ → 2OH• (initiation)
Step 2: OH• + H₂O₂ → HO₂• + H₂O (propagation)
Step 3: HO₂• + H₂O₂ → H₂O + O₂ + OH• (propagation)
b) Radical Mechanisms
- Often involve free radicals
- Chain reactions common
- Temperature sensitive
7. Factors Affecting Decomposition
a) Physical Factors
- Temperature
- Pressure
- Surface area
- Particle size
b) Chemical Factors
- Presence of catalysts
- pH of medium
- Presence of impurities
8. Industrial Applications
a) Metallurgy
- Ore decomposition
- Metal extraction
- Purification
b) Manufacturing
- Lime production
- Cement industry
- Fertilizer production
9. Advanced Problem Types
a) Stoichiometric Calculations
JEE Advanced Example (2020):
Question: Calculate volume of O₂ produced at STP when 12.6g of KClO₃ decomposes completely.
Solution:
- 2KClO₃ → 2KCl + 3O₂
- Moles of KClO₃ = 12.6/122.5 = 0.103
- Moles of O₂ = 0.103 × 3/2 = 0.154
- Volume = 0.154 × 22.4 = 3.45L
b) Mixed Problems
- Multiple reactions
- Competing decomposition paths
- Complex stoichiometry
10. Special Cases and Exceptions
a) Unusual Decompositions
- NH₄OH → NH₃ + H₂O
- NH₄HCO₃ → NH₃ + H₂O + CO₂
b) Catalytic Decomposition
Examples:
- H₂O₂ → H₂O + ½O₂ (MnO₂ catalyst)
- KClO₃ → KCl + 3/2O₂ (MnO₂ catalyst)
Important Points for JEE Advanced:
1. Remember stability orders
2. Know standard decomposition products
3. Practice numerical problems
4. Understand mechanism variations
5. Focus on exceptions
Common JEE Questions Types:
1. Stoichiometric calculations
2. Gas volume calculations
3. Thermal stability comparisons
4. Mechanism identification
5. Product prediction
Tips for Problem Solving:
1. Write balanced equations first
2. Check for catalysts
3. Consider reaction conditions
4. Look for competing reactions
5. Verify stoichiometry carefully
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Q 1. The reaction, \(2 \stackrel{+1}{\mathrm{H}_{2}}-\frac{2}{\mathrm{O}}(l) \xrightarrow{\Delta} 2 \stackrel{0}{\mathrm{H}_{2}}(g)+\stackrel{0}{\mathrm{O}_{2}}(g)\) is an example of; |
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(a) addition reaction; |
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(b) decomposition reaction; |
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(c) displacement reaction; |
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(d) None of these; |
REDUCTION REACTIONS
1. Basic Concepts
a) Definition and Types
- Removal of oxygen
- Addition of hydrogen
- Addition of electropositive element
- Gain of electrons
- Decrease in oxidation state
b) Examples for each type:
- CuO + H₂ → Cu + H₂O (Oxygen removal)
- CH₂=CH₂ + H₂ → CH₃-CH₃ (Hydrogen addition)
- 2Na⁺ + 2e⁻ → 2Na (Electron gain)
2. Types of Reducing Agents
a) Elemental Reducing Agents
- Hydrogen (H₂)
- Carbon (C)
- Active metals (Na, Mg, Al, etc.)
JEE Advanced Example (2019):
Question: Arrange in order of reducing power:
CO, H₂, C
Solution:
At high temperature: C > CO > H₂
At low temperature: H₂ > CO > C
b) Compound Reducing Agents
- Carbon monoxide (CO)
- Hydrogen sulfide (H₂S)
- Sodium borohydride (NaBH₄)
- Lithium aluminium hydride (LiAlH₄)
3. Industrial Applications
a) Metallurgy
Examples:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
JEE Advanced Example (2018):
Question: In the extraction of iron from its oxide, the reducing agent is:
Solution:
- Primary: CO (from coke)
- Secondary: C (direct reduction)
Equation: Fe₂O₃ + 3CO → 2Fe + 3CO₂
b) Organic Synthesis
- Hydrogenation of oils
- Reduction of functional groups
- Synthesis of alcohols
4. Electrochemical Aspects
a) Standard Reduction Potentials
- More positive E° = Better oxidizing agent
- More negative E° = Better reducing agent
JEE Advanced Example (2017):
Question: Calculate ΔG° for:
Zn + Cu²⁺ → Zn²⁺ + Cu
Given: E°(Cu²⁺/Cu) = +0.34V
E°(Zn²⁺/Zn) = -0.76V
Solution:
ΔG° = -nFE°
E° = 0.34 - (-0.76) = 1.10V
ΔG° = -2 × 96500 × 1.10 = -212.3 kJ/mol
b) Cell Reactions
- Cathode = Reduction
- E°cell = E°reduction - E°oxidation
5. Mechanisms of Reduction
a) Direct Electron Transfer
- Metal ion reduction
- Electrochemical processes
b) Hydride Transfer
- Organic reductions
- Complex metal hydrides
c) Hydrogen Transfer
- Catalytic hydrogenation
- Metal-acid reactions
6. Important Industrial Reducing Agents
a) Carbon and CO
- Used in metallurgy
- High temperature reactions
- Blast furnace operations
b) Hydrogen
- Clean reducing agent
- Catalytic reactions
- Organic synthesis
7. Selective Reduction
a) Organic Compounds
JEE Advanced Example (2020):
Question: Predict products when CH₃-CO-CH₃ is reduced with:
(i) LiAlH₄
(ii) NaBH₄
Solution:
Both give CH₃-CHOH-CH₃
LiAlH₄ is stronger but both can reduce ketones
b) Inorganic Compounds
- Metal ion reductions
- Selective precipitation
8. Thermodynamic Aspects
a) Energy Changes
- Usually exothermic
- ΔH negative
- Spontaneity depends on ΔG
b) Entropy Changes
- Generally decreases
- Important in gas phase reactions
- Temperature dependent
9. Kinetic Considerations
a) Rate Determining Steps
- Electron transfer
- Bond breaking/formation
- Mass transfer
b) Catalysis
- Heterogeneous catalysts (Pt, Pd, Ni)
- Homogeneous catalysts
- Biological reduction
10. Special Cases and Applications
a) Complex Reductions
JEE Advanced Example (2016):
Question: In the reduction of KMnO₄, identify products in:
(i) Acidic medium
(ii) Basic medium
Solution:
Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Basic: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
b) Biological Reduction
- NADH/NADPH systems
- Electron transport chain
- Photosynthesis
11. Important Points for JEE
a) Common Questions Types
- Electrode potential calculations
- Product prediction
- Mechanism identification
- Reducing agent comparison
b) Problem-Solving Strategy
- Write half-reactions
- Calculate E° values
- Consider reaction conditions
- Check stoichiometry
12. Practical Applications
a) Laboratory Methods
- Synthesis
- Analysis
- Purification
b) Industrial Processes
- Metal extraction
- Organic synthesis
- Food industry
Important Tips for JEE Advanced:
1. Remember activity series
2. Know standard reduction potentials
3. Understand selective reduction
4. Master electrochemical calculations
5. Focus on industrial applications
Common Mistakes to Avoid:
1. Ignoring pH conditions
2. Wrong stoichiometry
3. Incomplete reactions
4. Overlooking side reactions
5. Incorrect E° calculations
Practice Focus Areas:
1. Numerical problems on E°
2. Product prediction
3. Mechanism writing
4. Industrial processes
5. Selective reduction
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Q 1. The number of electrons involved in the reduction of one nitrate ion to hydrazine is; |
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(a) 8; |
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(b) 5; |
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(c) 3; |
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(d) 7; |
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Q 2. Arrange the following in the order of their decreasing electrode potentials : \(\mathrm{Mg}, \mathrm{K}, \mathrm{Ba}\) and \(\mathrm{Ca}\); |
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(a) \(\mathrm{K}, \mathrm{Ca}, \mathrm{Ba}, \mathrm{Mg}\); |
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(b) \(\mathrm{Ba}, \mathrm{Ca}, \mathrm{K}, \mathrm{Mg}\); |
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(c) \(\mathrm{Ca}, \mathrm{Mg}, \mathrm{K}, \mathrm{Ba}\); |
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(d) \(\mathrm{Mg}, \mathrm{Ca}, \mathrm{Ba}, \mathrm{K}\); |
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Q 3. If aqueous solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is made acidic For this which of the following statement(s) is/are correct? (i) This aqueous solution oxidizes \(\mathrm{I}^{-}\) (ii) This aqueous solution oxidizes \(\mathrm{F}^{-}\); |
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(a) Both statements (i) and (ii) are correct.; |
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(b) Statement (i) is correct and (ii) is incorrect.; |
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(c) Statement (ii) is correct and (i) is incorrect.; |
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(d) Both statements (i) and (ii) are incorrect.; |
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Q 4. Standard reduction potentials of the half-reactions are given below: \[ \begin{align*} &\mathrm{F}_{2}(g) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{~F}^{-}(aq) & \mathrm{E}^{\circ} &= +2.85 \mathrm{~V} \\ &\mathrm{Cl}_{2}(g) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(aq) & \mathrm{E}^{\circ} &= +1.36 \mathrm{~V} \\ &\mathrm{Br}_{2}(l) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(aq) & \mathrm{E}^{\circ} &= +1.06 \mathrm{~V} \\ &\mathrm{I}_{2}(s) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-}(aq) & \mathrm{E}^{\circ} &= +0.53 \mathrm{~V} \\ \end{align*} \] The strongest oxidizing agent and reducing agent respectively are: ; |
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(a) \(\mathrm{F}_{2}\) and \(\mathrm{I}^{-}\); |
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(b) \(\mathrm{Br}_{2}\) and \(\mathrm{Cl}^{-}\); |
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(c) \(\mathrm{Cl}_{2}\) and \(\mathrm{Br}^{-}\); |
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(d) \(\mathrm{Cl}_{2}\) and \(\mathrm{I}_{2}\); |
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Q 5. Standard electrode potentials of redox couples \(\mathrm{A}^{2+} / \mathrm{A}, \mathrm{B}^{2+} / \mathrm{B}, \mathrm{C} / \mathrm{C}^{2+}\) and \(\mathrm{D}^{2+} / \mathrm{D}\) are \(0.3 \mathrm{~V},-0.5 \mathrm{~V},-0.75 \mathrm{~V}\) and \(0.9 \mathrm{~V}\) respectively. Which of these is best oxidising agent and reducing agent respectively -; |
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(a) \(\mathrm{D}^{2+} / \mathrm{D}^{2}\) and \(\mathrm{B}^{2+} / \mathrm{B}\); |
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(b) \(\mathrm{B}^{2+} / \mathrm{B}\) and \(\mathrm{D}^{2+} / \mathrm{D}\); |
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(c) \(\mathrm{D}^{2+} / \mathrm{D}\) and \(\mathrm{C}^{2+} / \mathrm{C}\); |
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(d) \(\mathrm{C}^{2+} / \mathrm{C}\) and \(\mathrm{D}^{2+} / \mathrm{D}\); |
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Q 6. The standard reduction potentials for \(\mathrm{Cu}^{2+} / \mathrm{Cu}\), \(\mathrm{Zn}^{2+} / \mathrm{Zn}\), \(\mathrm{Li}^{+} / \mathrm{Li}\), \(\mathrm{Ag}^{+} / \mathrm{Ag}\), and \(\mathrm{H}^{+} / \mathrm{H}_{2}\) are \(+0.34 \mathrm{~V}, -0.762 \mathrm{~V}, -3.05 \mathrm{~V}, +0.80 \mathrm{~V},\) and \(0.00 \mathrm{~V}\) respectively. Choose the strongest reducing agent among the following: ; |
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(a) \(\mathrm{Zn}\); |
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(b) \(\mathrm{H}_{2}\); |
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(c) \(\mathrm{Ag}\); |
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(d) \(\mathrm{Li}\); |
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Q 7. Given: \[ \begin{aligned} & \mathrm{E}_{\frac{1}{2} \mathrm{Cl}_{2} / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36 \mathrm{~V}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\mathrm{o}}=-0.74 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} / \mathrm{Cr}^{3+}}^{\mathrm{o}}=1.33 \mathrm{~V}, \mathrm{E}_{\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}}^{\mathrm{o}}=1.51 \mathrm{~V} \end{aligned} \] The correct order of reducing power of the species \(\left(\mathrm{Cr}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}\right.\) and \(\mathrm{Cl}^{-}\)) will be; |
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(a) \(\mathrm{Mn}^{2+}<\mathrm{Cl}^{-}<\mathrm{Cr}^{3+}<\mathrm{Cr}\); |
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(b) \(\mathrm{Mn}^{2+}<\mathrm{Cl}^{3+}<\mathrm{Cl}^{-}<\mathrm{Cr}\); |
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(c) \(\mathrm{Cr}^{3+}<\mathrm{Cl}^{-}<\mathrm{Mn}^{2+}<\mathrm{Cr}\); |
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(d) \(\mathrm{Cr}^{3+}<\mathrm{Cl}^{-}<\mathrm{Cr}<\mathrm{Mn}^{2+}\); |
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Q 8. Which of the following statements are correct concerning redox properties? (i) A metal \(\mathrm{M}\) for which \(\mathrm{E}^{\circ}\) for the half life reaction \(\mathrm{M}^{\mathrm{n}+}+\mathrm{ne}^{-} \rightleftharpoons \mathrm{M}\) is very negative will be a good reducing agent. (ii) The oxidizing power of the halogens decreases from chlorine to iodine. (iii) The reducing power of hydrogen halides increases from hydrogen chloride to hydrogen iodide; |
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(a) (i), (ii) and(iii); |
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(b) (i) and (ii); |
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(c) (i) only; |
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(d) (ii) and (iii); |
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Q 9. The standard electrode potential \(\left(\mathrm{E}^{\circ}\right)\) values of \(\mathrm{Al}^{3+} / \mathrm{Al}, \mathrm{Ag}^{+} / \mathrm{Ag}, \mathrm{K}^{+} / \mathrm{K}\) and \(\mathrm{Cr}^{3+} / \mathrm{Cr}\) are \(-1.66 \mathrm{~V}\), \(0.80 \mathrm{~V},-2.93 \mathrm{~V}\) and \(-0.74 \mathrm{~V}\), respectively. The correct decreasing order of reducing power of the metal is [NEET Odisha 2019]; |
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(a) \(\mathrm{Al}>\mathrm{K}>\mathrm{Ag}>\mathrm{Cr}\); |
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(b) \(\mathrm{Ag}>\mathrm{Cr}>\mathrm{Al}>\mathrm{K}\); |
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(c) \(\mathrm{K}>\mathrm{Al}>\mathrm{Cr}>\mathrm{Ag}\); |
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(d) \(\mathrm{K}>\mathrm{Al}>\mathrm{Ag}>\mathrm{Cr}\); |
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Q 10. Standard reduction potentials of the half reactions are given below : \(\mathrm{F}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{~F}^{-}(\mathrm{aq}) ; E^{\circ}=+2.85 \mathrm{~V}\) \(\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq}) ; E^{\circ}=+1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}(\mathrm{l})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(\mathrm{aq}) ; E^{\circ}=+1.06 \mathrm{~V}\) \(\mathrm{I}_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) ; E^{\circ}=+0.53 \mathrm{~V}\) The strongest oxidising and reducing agents respectively are : [2012 M]; |
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(a) \(\mathrm{F}_{2}\) and \(\mathrm{I}^{-}\); |
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(b) \(\mathrm{Br}_{2}\) and \(\mathrm{Cl}^{-}\); |
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(c) \(\mathrm{Cl}_{2}\) and \(\mathrm{Br}^{-}\); |
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(d) \(\mathrm{Cl}_{2}^{2}\) and \(\mathrm{I}_{2}\); |
BALANCING REDOX REACTIONS
1. Basic Concepts
a) Principles
- Conservation of atoms
- Conservation of charge
- Change in oxidation states
- Electron transfer balance
b) Prerequisites
- Oxidation state rules
- Identification of redox pairs
- Understanding of half-reactions
2. Methods of Balancing
a) Oxidation Number Method
Steps:
1. Assign oxidation states
2. Identify species being oxidized/reduced
3. Calculate electron change
4. Balance electrons by cross multiplication
5. Balance remaining atoms
JEE Advanced Example (2019):
Question: Balance using oxidation number method:
MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic medium)
Solution:
1. Oxidation states: Mn(+7 → +2), Fe(+2 → +3)
2. Electron change: Mn(+7 → +2) = 5e⁻, Fe(+2 → +3) = -1e⁻
3. Balance: 5Fe²⁺ needed for each MnO₄⁻
4. Final: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
b) Half-Reaction Method
Steps:
1. Write separate half-reactions
2. Balance atoms except H and O
3. Balance O by adding H₂O
4. Balance H by adding H⁺
5. Balance charges with electrons
6. Multiply to equalize electrons
7. Add half-reactions
3. Medium-Specific Balancing
a) Acidic Medium
JEE Advanced Example (2018):
Question: Balance in acidic medium:
Cr₂O₇²⁻ + SO₂ → Cr³⁺ + SO₄²⁻
Solution:
Oxidation: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Final: Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O
b) Basic Medium
Special considerations:
- Add OH⁻ instead of H⁺
- Convert H⁺ to H₂O by adding OH⁻
- Convert excess H₂O to OH⁻
4. Complex Cases
a) Multiple Element Changes
Example: KMnO₄ + KI → MnO₂ + I₂
JEE Advanced Example (2017):
Question: Balance:
CN⁻ + MnO₄⁻ → CNO⁻ + MnO₂
Solution:
2MnO₄⁻ + 3CN⁻ + H₂O → 3CNO⁻ + 2MnO₂ + 2OH⁻
b) Disproportionation Reactions
Example: Cl₂ + OH⁻ → Cl⁻ + ClO⁻
5. Common Mistakes to Avoid
a) Technical Errors
- Missing H⁺/OH⁻ ions
- Incorrect electron count
- Wrong oxidation states
b) Conceptual Errors
- Ignoring medium (acid/base)
- Missing spectator ions
- Incomplete balancing
6. Advanced Applications
a) Organic Redox Reactions
JEE Advanced Example (2016):
Question: Balance:
C₂H₅OH + Cr₂O₇²⁻ → CH₃COOH + Cr³⁺
Solution:
3C₂H₅OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O
b) Complex Ion Reactions
Example: [Fe(CN)₆]⁴⁻ + MnO₄⁻ → [Fe(CN)₆]³⁻ + MnO₂
7. Special Cases
a) Auto-oxidation Reactions
Example: 2CuCl → Cu + CuCl₂
b) Comproportionation
Example: Cl₂ + Cl⁻ → Cl₃⁻
8. Problem-Solving Strategy
a) General Approach
1. Identify type of reaction
2. Choose appropriate method
3. Follow systematic steps
4. Verify final equation
b) Verification Steps
- Atom balance
- Charge balance
- Oxidation state changes
9. Common JEE Questions
a) Direct Balancing
JEE Advanced Example (2020):
Question: Balance and find ratio of reactants:
MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂
Solution:
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
Ratio MnO₄⁻:C₂O₄²⁻ = 2:5
b) Product Prediction
- Given reactants
- Medium specified
- Multiple possible products
10. Important Tips for JEE
a) Preparation Strategy
- Practice standard reactions
- Memorize common redox pairs
- Know oxidation states
b) Exam Techniques
- Read conditions carefully
- Choose appropriate method
- Verify final answer
11. Practice Focus Areas
a) Numerical Problems
- Stoichiometric calculations
- Equivalent weight problems
- Concentration-based questions
b) Conceptual Questions
- Method selection
- Medium identification
- Product prediction
12. Advanced Level Concepts
a) Electrochemical Aspects
- Cell reactions
- Standard potentials
- Spontaneity
b) Industrial Applications
- Metallurgical processes
- Chemical manufacturing
- Analytical procedures
Key Points to Remember:
1. Always identify oxidizing/reducing agents first
2. Consider the medium (acidic/basic)
3. Follow systematic approach
4. Verify balancing at each step
5. Check final equation thoroughly
Common JEE Question Types:
1. Complete balancing
2. Finding reactant ratios
3. Product prediction
4. Medium-specific reactions
5. Complex ion balancing
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Q 1. Consider the following reaction : \[ \mathrm{xMNO}_{4}^{-}+\mathrm{yC}_{2} \mathrm{O}_{4}{ }^{2-}+\mathrm{zH}^{+} \rightarrow \mathrm{xMn}^{2+}+2 \mathrm{yCO}_{2}+\frac{\mathrm{z}}{2} \mathrm{H}_{2} \mathrm{O} \] The value's of \(x\), \(y\) and \(z\) in the reaction are, respectively :; |
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(a) 5, 2 and 16; |
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(b) 2, 5 and 8; |
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(c) 2, 5 and 16; |
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(d) 5,2 and 8; |
Standard Electrode Potential
1. Basic Concepts
- Definition: Standard electrode potential (E°) is the potential of an electrode when the concentration of ions in solution is 1 M, pressure is 1 atm, and temperature is 298 K.
- Principles:
- Electrode Potential: Measures the tendency of a species to gain or lose electrons.
- Reduction and Oxidation Potentials: E° values are usually given for reduction reactions.
- Reference Electrode: The standard hydrogen electrode (SHE) is used as a reference with E° = 0 V.
2. Calculation of Electrode Potential
- Electrode Potential Equation:
- \( E = E° + \frac{0.0591}{n} \log \frac{[Ox]}{[Red]} \) (at 298 K)
- Where \( n \) = number of electrons transferred.
- Nernst Equation:
- For a reaction: \( M^z + z e^- \rightarrow M \),
- \( E = E° - \frac{0.0591}{z} \log [M^z] \).
3. Determining Standard Electrode Potential (E°)
a) Direct Measurement with SHE:
- The SHE is used as a reference (0 V), and E° of other electrodes is measured relative to it.
b) Using Known Reactions:
- Standard potentials are assigned to half-cells by measuring voltage against a known half-cell.
4. Applications of Standard Electrode Potentials
a) Predicting Redox Reactions:
- Reaction Feasibility: A species with higher E° (more positive) acts as a stronger oxidizing agent.
- Example: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) (Cu²⁺ has a higher E°, so it can oxidize Zn).
b) Electrochemical Series:
- Lists elements based on E° values, helping predict the direction of electron flow.
- Important Points: Metals with low E° are better reducing agents; non-metals with high E° are good oxidizing agents.
5. Calculation of Cell Potential
a) Galvanic Cell Potential (E°cell):
- \( E°_{cell} = E°_{cathode} - E°_{anode} \).
b) Example Problem:
- Calculate \( E°_{cell} \) for the cell Zn | Zn²⁺ || Cu²⁺ | Cu.
- Solution: \( E°_{cell} = E°_{Cu^{2+}/Cu} - E°_{Zn^{2+}/Zn} \).
6. Medium-Specific Considerations
- Acidic Medium: Often changes E° values due to interaction with H⁺ ions.
- Basic Medium: Adds OH⁻ ions, affecting the redox potential and equilibrium.
7. Significance in JEE Advanced
a) Electrochemical Cell Questions:
- Calculation of cell potential using standard electrode potentials.
- Example: Calculate E° for the cell Fe³⁺/Fe²⁺ || Cu²⁺/Cu.
b) Redox Predictions:
- Use E° values to determine spontaneous reactions in redox systems.
8. Important Trends in Standard Electrode Potential
- Metals:
- Alkali metals have low E° values, acting as strong reducing agents.
- Noble metals like Au have high E° values, making them stable and less reactive.
- Non-Metals:
- Halogens have high E° values, making them strong oxidizing agents.
9. Advanced Applications
a) Battery Chemistry:
- Understanding how cells like lead-acid, dry cell, and lithium-ion batteries operate.
b) Industrial Applications:
- Electrode potentials are crucial in electroplating, corrosion protection, and electrorefining.
10. Problem-Solving Strategy for JEE
a) General Steps:
1. Identify the half-reactions.
2. Use the electrochemical series to predict cell reactions.
3. Apply the Nernst equation if conditions deviate from standard.
b) Verification Steps:
- Ensure charge and mass balance.
- Cross-check with known E° values.
11. Practice Focus Areas
a) Numerical Problems:
- Calculating cell potential under non-standard conditions.
- Stoichiometry and concentration effects.
b) Conceptual Questions:
- Predicting spontaneity.
- Relating E° values to reaction feasibility.
12. Advanced Level Concepts
a) Thermodynamic Aspects:
- Relation to Gibbs free energy: \( \Delta G° = -nFE° \).
b) Electrolysis and Faraday’s Laws:
- Quantitative aspects of electrolysis and electrodeposition.
Key Points to Remember:
1. Higher E° means stronger oxidizing agent.
2. E° values are relative to the SHE.
3. Use Nernst equation for non-standard conditions.
4. Cross-reference with the electrochemical series.
5. Verify charge and mass balance in calculations.
Common JEE Question Types:
1. Calculating cell potential.
2. Predicting redox reactions.
3. Application of Nernst equation.
4. Relation of E° with spontaneity.
5. Electrochemical series-based comparisons.
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Q 1. Consider the following standard electrode potentials ( \(E^{\circ}\) in volts) in aqueous solution: Element & \(\mathbf{M}^{3+} / \mathbf{M}\) & \(\mathbf{M}^{+} / \mathbf{M}\) \\ \(\mathrm{Al}\) & -1.66 & +0.55 \\ \(\mathrm{Tl}\) & +1.26 & -0.34 Based on these data, which of the following statements is correct?; |
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(a) \(\mathrm{Tl}^{+}\)is more stable than \(\mathrm{Al}^{+}\); |
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(b) \(\mathrm{Tl}^{3+}\) is more stable than \(\mathrm{Al}^{3+}\); |
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(c) \(\mathrm{Al}^{+}\)is more stable than \(\mathrm{Al}^{3+}\); |
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(d) \(\mathrm{Tl}^{+}\)is more stable than \(\mathrm{Al}^{3+}\); |
General Redox Reaction Concepts |
Oxidation Reaction |
Reducing Agent in Redox Reaction |
Redox Reaction |
Displacement Reaction - Redox reaction where one element replaces another in a compound. |
Disproportionation Reaction - Reaction where an element undergoes both oxidation and reduction. |
Decomposition Reaction - Redox reaction breaking a compound into simpler components. |
Reduction - Process involving removal of oxygen or addition of hydrogen/electropositive element. |
Balancing Redox Reactions - Techniques include oxidation number and half-reaction methods. |
Standard Electrode Potential - Potential of an electrode under standard conditions. |