chapter 5 electrochemistry class 12 chemistry textbook solution

3. Answer the following in brief

iii. Write electrode reactions for the electrolysis of aqueous NaCl.


1. Reduction half-reaction at cathode: - At the cathode, two reduction reactions compete: i. Reduction of sodium ions:
\[Na_{(aq)}^+ + e^- \rightarrow Na_{(s)}, \quad E^\circ = -2.71 \, \text{V}\]
ii. Reduction of water to hydrogen gas:
\[2H_2O_{(l)} + 2e^- \rightarrow H_2(g) + 2OH_{(aq)}^-, \quad E^\circ = -0.83 \, \text{V}\]
- The standard potential for the reduction of water is higher than that for the reduction of Na⁺. Therefore, water has a greater tendency to get reduced than the Na⁺ ion. Hence, reduction of water is the cathode reaction during the electrolysis of aqueous NaCl. 2. Oxidation half-reaction at anode: - At the anode, there will be competition between the oxidation of Cl⁻ ions to chlorine gas (as in the case of molten NaCl) and the oxidation of water to O₂ gas: i. Oxidation of Cl⁻ ions to chlorine gas:
\[2Cl_{(aq)}^- \rightarrow Cl_{2(g)} + 2e^-, \quad E^\circ_{oxd} = -1.36 \, \text{V}\]
ii. Oxidation of water to oxygen gas:
\[2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H_{(aq)}^+ + 2e^-, \quad E^\circ_{oxd} = -0.4 \, \text{V}\]
- The standard electrode potential for the oxidation of water is greater than that of Cl⁻ ions, indicating that water has a greater tendency to undergo oxidation. However, experiments have shown that the gas produced at the anode is Cl₂ and not O₂. This suggests that the anode reaction is the oxidation of Cl⁻ to Cl₂ gas, likely due to overvoltage. 3. Net cell reaction: - The net cell reaction is the sum of the two electrode reactions:
\[2Cl_{(aq)}^- \rightarrow Cl_{2(g)} + 2e^- \quad \text{(Oxidation half-reaction at anode)}\] \[2H_{2}O_{(l)} + 2e^- \rightarrow H_{2(g)} + 2OH_{(aq)}^- \quad \text{(Reduction half-reaction at cathode)}\]
- The overall cell reaction is given by:
\[2Cl_{(aq)}^- + 2H_{2}O_{(l)} \rightarrow Cl_{2(g)} + H_{2(g)} + 2OH_{(aq)}^- \quad \text{(Overall cell reaction)}\]

chapter 5 elctrochemistry textbook solution page 118