If mercury is used as a cathode during the electrolysis of an aqueous solution, the ions discharged — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Ions in the Aqueous Solution An aqueous solution of sodium chloride ($\ce{NaCl}$) contains dissolved sodium ions ($\ce{Na^+}$) and chloride ions ($\ce{Cl^-}$). Additionally, the autoionization of water molecules contributes small concentrations of hydrogen ions ($\ce{H^+}$) and hydroxide ions ($\ce{OH^-}$): $$\ce{NaCl(aq) -> Na^+(aq) + Cl^-(aq)}$$ $$\ce{H2O(l) <=> H^+(aq) + OH^-(aq)}$$ Step 2 - Analyze Cathodic Competition on Inert Electrodes The cathode is the negative electrode where reduction (gain of electrons) occurs. Cations ($\ce{Na^+}$ and $\ce{H^+}$) migrate toward the cathode. The competing reduction half-reactions and their standard reduction potentials ($E^\circ$) at $25^\circ\text{C}$ are: 1. **Reduction of Hydrogen Ions (or water):** $$\ce{2H^+(aq) + 2e^- -> H2(g)} \quad E^\circ = 0.00\text{ V}$$ $$\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)} \quad E^\circ = -0.83\text{ V}$$ 2. **Reduction of Sodium Ions:** $$\ce{Na^+(aq) + e^- -> Na(s)} \quad E^\circ = -2.71\text{ V}$$ Under standard conditions with inert electrodes (such as platinum or carbon), $\ce{H^+}$ ions (or water) are preferentially reduced because their standard reduction potential is significantly more positive (less negative) than that of $\ce{Na^+}$ ($0.00\text{ V} > -2.71\text{ V}$). Consequently, hydrogen gas ($\ce{H2}$) is evolved, and sodium ions remain unchanged in the solution. Step 3 - Understand the Influence of the Mercury Cathode When liquid mercury ($\ce{Hg}$) is used as the cathode, the preferential discharge order is completely reversed due to two major factors: 1. **High Overvoltage of Hydrogen on Mercury (Kinetic Factor):** Overvoltage is the additional potential beyond the thermodynamic standard potential required to overcome kinetic barriers at the electrode interface. The activation energy barrier for the reduction of $\ce{H^+}$ to form $\ce{H2(g)}$ on a liquid mercury surface is exceptionally high. This kinetic overpotential ($> 1.0\text{ V}$) shifts the actual discharge potential of hydrogen to a highly negative, unfavorable value. 2. **Sodium Amalgam Formation (Thermodynamic Factor):** Sodium metal has an outstanding thermodynamic affinity for mercury and dissolves in it spontaneously to form a stable liquid/solid solution called sodium amalgam ($\ce{Na-Hg}$). This amalgamation is highly exothermic: $$\ce{Na^+(aq) + e^- + Hg(l) -> Na-Hg} \quad \Delta H < 0$$ The formation of sodium amalgam decreases the chemical activity of the reduced sodium metal, which shifts the reduction potential of the sodium ion to a much less negative (more favorable) value. Step 4 - Write the Cathodic Reduction Half-Reaction Due to the combined kinetic inhibition of hydrogen evolution and thermodynamic stabilization of sodium amalgam, the $\ce{Na^+}$ ions are preferentially discharged at the mercury cathode: $$\ce{Na^+(aq) + e^- + Hg(l) -> Na-Hg}$$ Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** Although $\ce{H^+}$ has a higher standard reduction potential, it is not discharged at a mercury cathode due to the extraordinarily high hydrogen overvoltage on mercury. * **Option (B) is correct:** $\ce{Na^+}$ ions are preferentially discharged to form sodium amalgam owing to the low activity of sodium in the amalgam phase and high hydrogen overpotential on mercury. * **Option (C) is incorrect:** $\ce{OH^-}$ is an anion, so it migrates to the anode (positive electrode) to undergo oxidation, not the cathode. * **Option (D) is incorrect:** $\ce{Cl^-}$ is an anion, so it migrates to the anode to undergo oxidation, not the cathode. $$\text{Correct Option: } \boxed{\text{B}}$$