During the electrolysis of an aqueous salt solution, the pH in the space near one of the electrode w — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Principles of Electrolysis in Aqueous Media During the electrolysis of an aqueous salt solution, several chemical species migrate toward the electrodes and compete to undergo electrochemical reactions: 1. **At the Cathode (Negative Electrode):** Cations from the salt and water molecules compete for reduction (gain of electrons). 2. **At the Anode (Positive Electrode):** Anions from the salt and water molecules compete for oxidation (loss of electrons). The relative ease of these reactions is governed by the standard reduction/oxidation potentials and the active concentrations of the species in the solution. Step 2 - Analyze the Electrolysis of Very Dilute Aqueous $\ce{NaCl}$ (Option A) In an extremely dilute aqueous solution of sodium chloride ($\ce{NaCl}$), the concentrations of sodium ions ($\ce{Na^+}$) and chloride ions ($\ce{Cl^-}$) are negligibly small, meaning that water molecules ($\ce{H2O}$) overwhelmingly dominate the electrochemical processes at both electrodes. * **Reaction at the Cathode (Reduction):** Both $\ce{Na^+(aq)}$ ions and $\ce{H2O(l)}$ molecules are present near the cathode: $$\ce{Na^+(aq) + e^- -> Na(s)} \quad E^\circ_{\text{red}} = -2.71\text{ V}$$ $$\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)} \quad E^\circ_{\text{red}} = -0.83\text{ V}$$ Since the reduction potential of water is significantly more positive (less negative) than that of sodium ions ($-0.83\text{ V} > -2.71\text{ V}$), water molecules are preferentially reduced. This produces hydrogen gas ($\ce{H2}$) and releases hydroxide ions ($\ce{OH^-}$) into the surrounding space: $$\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)}$$ The local accumulation of basic $\ce{OH^-}$ ions **increases the pH** ($\text{pH} > 7$) near the cathode. * **Reaction at the Anode (Oxidation):** Both $\ce{Cl^-(aq)}$ ions and $\ce{H2O(l)}$ molecules are present near the anode: $$\ce{2Cl^-(aq) -> Cl2(g) + 2e^-} \quad E^\circ_{\text{ox}} = -1.36\text{ V}$$ $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \quad E^\circ_{\text{ox}} = -1.23\text{ V}$$ Although chloride oxidation and water oxidation have comparable potentials, in a highly dilute solution, the concentration of $\ce{Cl^-}$ is too low to overcome the concentration barrier. Thus, water molecules are preferentially oxidized, releasing oxygen gas ($\ce{O2}$) and hydrogen ions ($\ce{H^+}$): $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}$$ The local accumulation of acidic $\ce{H^+}$ ions **decreases the pH** ($\text{pH} < 7$) near the anode. Thus, during the electrolysis of very dilute $\ce{NaCl(aq)}$, the pH increases near the cathode and decreases near the anode, satisfying the given condition. Step 3 - Analyze the Other Options * **Option (B) $\ce{ZnCl2}$:** * At the Cathode: Zinc ions ($\ce{Zn^2+}$) have a standard reduction potential of $-0.76\text{ V}$, which is more favorable than water reduction ($-0.83\text{ V}$). Thus, zinc metal is deposited ($\ce{Zn^2+ + 2e^- -> Zn}$), and no $\ce{OH^-}$ is formed, leaving the pH at the cathode unchanged. * At the Anode: Chloride ions are oxidized to chlorine gas ($\ce{2Cl^- -> Cl2 + 2e^-}$), and no $\ce{H^+}$ is formed. * **Option (C) Concentrated $\ce{NaCl}$:** * At the Cathode: Water is reduced to $\ce{H2}$ and $\ce{OH^-}$, increasing the pH. * At the Anode: Due to high chloride concentration and overvoltage factors, chloride ions are preferentially oxidized to form chlorine gas ($\ce{2Cl^- -> Cl2 + 2e^-}$). Since water is not oxidized, no $\ce{H^+}$ ions are produced, and the pH near the anode remains unchanged. * **Option (D) $\ce{Cu(NO3)2}$:** * At the Cathode: Copper ions ($\ce{Cu^2+}$, $E^\circ_{\text{red}} = +0.34\text{ V}$) are easily reduced to metallic copper ($\ce{Cu^2+ + 2e^- -> Cu}$), so the pH near the cathode remains unchanged. * At the Anode: Nitrate ions ($\ce{NO3^-}$) are highly stable and resistant to oxidation. Therefore, water is oxidized to form $\ce{O2}$ and $\ce{H^+}$, decreasing the pH near the anode. Step 4 - Conclusion Only the electrolysis of very dilute $\ce{NaCl}$ solution (Option A) causes water to be oxidized at the anode (decreasing pH) and reduced at the cathode (increasing pH) simultaneously. $$\text{Correct Option: } \boxed{\text{A}}$$