Statement I: During electrolysis of aqueous sodium acetate solution, the molar ratio of gases at cat — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the Dissociation of Aqueous Sodium Acetate Aqueous sodium acetate ($\ce{CH3COONa}$) is a strong electrolyte that dissociates completely in water to yield sodium ions ($\ce{Na^+}$) and acetate ions ($\ce{CH3COO^-}$): $$\ce{CH3COONa(aq) -> CH3COO^-(aq) + Na^+(aq)}$$ Additionally, water undergoes self-ionization to a small extent, providing hydrogen ions ($\ce{H^+}$) and hydroxide ions ($\ce{OH^-}$): $$\ce{H2O(l) <=> H^+(aq) + OH^-(aq)}$$ During electrolysis, competition occurs at both electrodes: * At the cathode, $\ce{Na^+(aq)}$ and $\ce{H^+(aq)}$ (or water molecules) compete for reduction. * At the anode, $\ce{CH3COO^-(aq)}$ and $\ce{OH^-(aq)}$ (or water molecules) compete for oxidation. Step 2 - Analyze Cathode Reactions and Hydrogen Evolution At the cathode, reduction of water or hydrogen ions occurs preferentially over sodium ions because the standard reduction potential of hydrogen/water is much higher than that of sodium ($E^\circ_{\ce{Na^+/Na}} = -2.71\text{ V}$): $$\ce{2H2O(l) + 2e^- -> H2(g) \uparrow + 2OH^-(aq)}$$ Alternatively, this can be written in terms of hydrogen ion discharge: $$\ce{2H^+(aq) + 2e^- -> H2(g) \uparrow}$$ Thus, for every $2\text{ moles}$ of electrons ($2\text{ F}$ of electricity) that pass through the cathode, exactly $1\text{ mole}$ of hydrogen gas ($\ce{H2}$) is evolved: $$n_{\text{gas, cathode}} = 1\text{ mol } \ce{H2}$$ Step 3 - Analyze Anode Reactions (Kolbe's Electrolysis) At the anode, acetate ions ($\ce{CH3COO^-}$) are preferentially oxidized over hydroxide ions or water molecules. This process is known as **Kolbe's electrolytic reaction**. The mechanism at the anode proceeds as follows: 1. **Discharge of acetate ion to form acetate free radical:** $$\ce{2CH3COO^-(aq) -> 2CH3COO^{\bullet} + 2e^-}$$ 2. **Decarboxylation of the acetate radical to form methyl radicals and carbon dioxide gas:** $$\ce{2CH3COO^{\bullet} -> 2CH3^{\bullet} + 2CO2(g) \uparrow}$$ 3. **Dimerization of methyl radicals to form ethane gas:** $$\ce{2CH3^{\bullet} -> C2H6(g) \uparrow}$$ Combining these steps gives the net balanced reaction at the anode: $$\ce{2CH3COO^-(aq) -> C2H6(g) \uparrow + 2CO2(g) \uparrow + 2e^-}$$ Thus, for every $2\text{ moles}$ of electrons ($2\text{ F}$ of electricity) lost at the anode, the total moles of evolved gases are: $$n_{\text{gas, anode}} = 1\text{ mol } \ce{C2H6} + 2\text{ mol } \ce{CO2} = 3\text{ mol of gas}$$ Step 4 - Calculate the Molar Ratio of Evolved Gases Now, we calculate the molar ratio of the total gases evolved at the cathode to those evolved at the anode during the same electrolytic period: $$\text{Molar Ratio} = \frac{\text{Moles of gas at cathode}}{\text{Moles of gas at anode}}$$ $$\text{Molar Ratio} = \frac{1\text{ mol }\ce{H2}}{3\text{ mol }(\ce{C2H6} + \ce{CO2})} = 1:3$$ Thus, the molar ratio of the gases evolved at the cathode and the anode is indeed $1:3$. Therefore, **Statement I is correct**. Step 5 - Evaluate both Statements and their Relationship * **Statement I is correct:** As calculated in Step 4, the molar ratio of gases evolved at the cathode and anode is $1:3$. * **Statement II is correct:** During the electrolysis, acetate ions ($\ce{CH3COO^-}$) are discharged at the anode to yield ethane and carbon dioxide, while hydrogen ions ($\ce{H^+}$) from water are discharged at the cathode to produce hydrogen gas. * **Analysis of the Relationship:** While Statement II correctly identifies *which* ionic species undergo discharge at each electrode, it does not provide the quantitative stoichiometric relationship (the balanced reactions) necessary to explain the $1:3$ molar ratio. The $1:3$ ratio is a direct consequence of the stoichiometry of the reduction and Kolbe's oxidation processes, not merely the identity of the discharging species. Therefore, both statements are correct, but Statement II is not the correct explanation of Statement I. $$\text{Correct Option: } \boxed{B}$$