JEE Main 2025 Apr 7 Shift 1, Chemistry Q23: Electrochemical Cell And Nernst Equation

Step 1: Determine post-electrolysis concentrations.

Cu${}^{2+}$/Cu cell: 1 F = 1 mol of electrons. Reduction Cu${}^{2+}$ + 2e${}^{-}$ $\to$ Cu consumes 0.5 mol of Cu${}^{2+}$ from the initial 1.5 mol in 1 L. Remaining: 1 mol in 1 L $\Rightarrow$ $[{\text{Cu}}^{2+}]=1$ M.

Ag${}^{+}$/Ag cell: 0.1 F deposits 0.1 mol of Ag, consuming 0.1 mol of Ag${}^{+}$ from the initial 0.2 mol. Remaining: 0.1 mol in 1 L $\Rightarrow$ $[{\text{Ag}}^{+}]=0.1$ M.

Step 2: Identify the spontaneous galvanic cell. Ag has the higher reduction potential, so Ag${}^{+}$ is reduced at the cathode and Cu is oxidised at the anode.

Net reaction: $\text{Cu}(s)+2{\text{Ag}}^{+}(aq)\to {\text{Cu}}^{2+}(aq)+2\text{Ag}(s)$, with $n=2$.

Step 3: Standard cell potential.

$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=0.80-0.34=0.46$ V

Step 4: Apply the Nernst equation.

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.06}{n}\log \frac{[{\text{Cu}}^{2+}]}{[{\text{Ag}}^{+}{]}^2}=0.46-\frac{0.06}{2}\log \frac{1}{(0.1{)}^2}$

$\log \frac{1}{0.01}=2$, so

$E_{\text{cell}}=0.46-0.03\times 2=0.46-0.06=0.40$ V $=400$ mV

$\therefore$ Answer = 400 mV.