Statement I: Sodium ions are discharged in preference to hydrogen ions at a mercury cathode. Stateme — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Evaluate Statement I (Discharge Preference of Cations on a Mercury Cathode) In the electrolysis of an aqueous solution of sodium chloride ($\ce{NaCl}$), there is competition at the cathode between the reduction of sodium ions ($\ce{Na^+}$) and hydrogen-determining species (water molecules or hydrogen ions, $\ce{H^+}$): * **Reduction of sodium ions:** $$\ce{Na^+(aq) + e^- -> Na(s)} \quad E^\circ = -2.71\text{ V}$$ * **Reduction of hydrogen ions / water:** $$\ce{2H^+(aq) + 2e^- -> H2(g)} \quad E^\circ = 0.00\text{ V}$$ $$\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)} \quad E^\circ = -0.83\text{ V}$$ Under standard conditions using inert electrodes (such as Platinum), $\ce{H2}$ gas is evolved at the cathode because its reduction potential is significantly more positive (higher) than that of sodium ions. However, when a liquid mercury ($\ce{Hg}$) cathode is employed, the relative discharge preference is completely reversed due to two major factors: 1. **High Overpotential of Hydrogen on Mercury:** The kinetic activation barrier (overpotential) for the evolution of hydrogen gas on a mercury metal surface is exceptionally high (ranging between $1.0\text{ V}$ and $1.5\text{ V}$). This effectively shifts the discharge potential of $\ce{H^+}$ to a much more negative value, making its reduction kinetically unfavorable. 2. **Thermodynamic Stabilization via Amalgam Formation:** Sodium metal, upon being reduced at the mercury cathode, immediately dissolves in the mercury to form a highly stable, low-activity solid/liquid solution called sodium amalgam ($\ce{Na-Hg}$): $$\ce{Na^+(aq) + e^- + Hg(l) -> Na(Hg)}$$ The formation of sodium amalgam is highly exothermic ($\Delta G^\circ \ll 0$), which thermodynamically shifts the effective reduction potential of sodium to a much less negative (more favorable) value. Consequently, because of the massive kinetic overpotential of hydrogen evolution on mercury and the thermodynamic assistance of amalgam formation, sodium ions ($\ce{Na^+}$) are discharged in preference to hydrogen ions ($\ce{H^+}$) at a mercury cathode. Therefore, **Statement I is correct**. Step 2 - Evaluate Statement II (Oxidizing and Reducing Nature of \ce{Na^+} and \ce{H^+} Ions) A **reducing agent** is defined as a species that loses/donates electrons and undergoes oxidation in a chemical reaction. In aqueous solution, both sodium ions ($\ce{Na^+}$) and hydrogen ions ($\ce{H^+}$) are present in their highest stable positive oxidation states. Under standard chemical conditions, they cannot undergo further oxidation (i.e., they cannot lose more electrons to form species like $\ce{Na^2+}$ or $\ce{H^2+}$). Instead, they can only accept electrons and undergo reduction, which makes them **oxidizing agents**, not reducing agents. If we compare their strengths as oxidizing agents based on their standard reduction potentials: $$E^\circ_{\ce{H^+/H2}} = 0.00\text{ V} > E^\circ_{\ce{Na^+/Na}} = -2.71\text{ V}$$ Since the reduction potential of $\ce{Na^+}$ is highly negative, it has a very weak tendency to gain electrons, which makes $\ce{Na^+}$ an extremely **weak oxidizing agent** compared to $\ce{H^+}$. *Note: While metallic sodium ($\ce{Na}$) is indeed a strong reducing agent, the sodium cation ($\ce{Na^+}$) itself is not.* Therefore, **Statement II is incorrect**. Step 3 - Determine the Correct Option Since Statement I is correct and Statement II is incorrect, the correct choice is Option (C). $$\text{Correct Option: } \boxed{C}$$