Suppose the concentration of hydroxide ion in the cell is doubled at 298 K. The cell voltage will be — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Half-Cell Reactions and the Net Cell Reaction A hydrogen-oxygen (\ce{H2-O2}) fuel cell operates in an alkaline medium. The half-cell reactions taking place at the electrodes are: * **Anode Half-Reaction (Oxidation):** $$\ce{2H2(g) + 4OH^-(aq) -> 4H2O(l) + 4e^-}$$ * **Cathode Half-Reaction (Reduction):** $$\ce{O2(g) + 2H2O(l) + 4e^- -> 4OH^-(aq)}$$ To obtain the net overall cell reaction, we add the two half-reactions together. Notice that the $4\text{ moles}$ of hydroxide ions ($\ce{OH^-}$) consumed at the anode are regenerated in equal amount ($4\text{ moles}$) at the cathode: $$\ce{2H2(g) + 4OH^-(aq) + O2(g) + 2H2O(l) + 4e^- -> 4H2O(l) + 4e^- + 4OH^-(aq)}$$ Canceling the common species ($\ce{OH^-}$, $\ce{e^-}$, and $\ce{H2O}$) on both sides yields the net overall reaction: $$\ce{2H2(g) + O2(g) -> 2H2O(l)}$$ Step 2 - Apply the Nernst Equation The Nernst equation relates the non-standard cell voltage ($E_{\text{cell}}$) to the standard cell voltage ($E^\circ_{\text{cell}}$) at $298\text{ K}$: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} Q$$ Where: * $n = 4$ (moles of electrons transferred in the balanced cell reaction). * $Q$ is the reaction quotient for the overall cell reaction. Step 3 - Set Up the Reaction Quotient ($Q$) The reaction quotient ($Q$) is formulated using the activities of the species involved in the overall balanced reaction: $$\ce{2H2(g) + O2(g) -> 2H2O(l)}$$ Since water is in the pure liquid phase, its activity is taken as unity ($a_{\ce{H2O}} = 1$). The activities of the gases are represented by their respective partial pressures ($P_{\ce{H2}}$ and $P_{\ce{O2}}$): $$Q = \frac{1}{(P_{\ce{H2}})^2 (P_{\ce{O2}})}$$ Crucially, the concentration of hydroxide ions ($[\ce{OH^-}]$) does not appear in the final expression for the reaction quotient $Q$ of the net cell reaction because it cancels out during the summation of the anode and cathode half-reactions. Step 4 - Determine the Effect of Doubling $[\ce{OH^-}]$ on the Cell Potential Since the concentration of hydroxide ions is absent from the net overall cell equation and the reaction quotient: $$\frac{\partial E_{\text{cell}}}{\partial [\ce{OH^-}]} = 0$$ Therefore, doubling the concentration of hydroxide ions in the electrolyte solution has absolutely no thermodynamic effect on the cell potential. The cell voltage remains completely unchanged. Step 5 - Evaluate the Options * **Option (A) is incorrect:** The voltage is not reduced by half because hydroxide concentration does not affect the net cell potential. * **Option (B) is incorrect:** The voltage is not doubled because the reaction quotient remains unaffected by the concentration of hydroxide ions. * **Option (C) is incorrect:** The voltage is not increased by a factor of 4. * **Option (D) is correct:** As mathematically demonstrated, the cell potential is independent of the hydroxide ion concentration. $$\text{Correct Option: } \boxed{D}$$