2025 H2 Chem P1 Q26 — Electrochemical cell Solution

The answer is C (silver half-cell less than +0.8 V; cell potential greater than +0.56 V).
From the Data Booklet: \(E^\ominus(\text{Cl}_2/\text{Cl}^-) = +1.36\) V and \(E^\ominus(\text{Ag}^+/\text{Ag}) = +0.80\) V. The chlorine half-cell is the positive electrode (reduction) and the silver half-cell is the negative electrode (oxidation):
\(E_\text{cell} = E(\text{positive electrode}) - E(\text{negative electrode}) = +1.36 - (+0.80) = +0.56\) V
Adding NaCl(aq) to the silver half-cell precipitates silver chloride: \(\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}\). \([\text{Ag}^+]\) falls, so the equilibrium \(\text{Ag}^+ + \text{e}^- \rightleftharpoons \text{Ag}\) shifts to the left. Reduction of \(\text{Ag}^+\) becomes less favourable, so the electrode potential of the silver half-cell becomes less than +0.80 V.
The silver half-cell is the negative electrode, so a less positive \(E(\text{Ag}^+/\text{Ag})\) is subtracted in \(E_\text{cell} = E(\text{Cl}_2/\text{Cl}^-) - E(\text{Ag}^+/\text{Ag})\), making \(E_\text{cell}\) greater than +0.56 V.
• A is wrong: removing \(\text{Ag}^+\) does change the potential of the silver half-cell, so neither value stays the same.
• B is wrong: the silver potential does fall, but because silver is the negative electrode this increases the cell potential, not decreases it.
• D is wrong: lowering \([\text{Ag}^+]\) makes the silver potential less positive, not greater than +0.8 V, and the cell potential increases.