Q.Calculate standard Gibbs energy change and equilibrium constant at 25$^0$C for the cell reaction, Cd (s) + Sn$^{2+}$ (aq) $\longrightarrow$ Cd$^{2+}$ (aq) + Sn (s) Given : $E^0_{Cd}$ = -0.403V and $E^0_{Sn}$ = - 0.136 V. Write formula of the cell.

Cd is oxidised (lower $E^0$) and Sn$^{2+}$ reduced, so $E^0_{cell}$ = -0.136 - (-0.403) = 0.267 V with $n$ = 2; then $\Delta G^0 = -nFE^0_{cell}$ and $\log_{10}K = nE^0_{cell}/0.0592$. …

Step 1 (cell formula). In the given reaction Cd is oxidised (anode) and Sn$^{2+}$ is reduced (cathode) -- consistent with $E^0_{Sn}$ (-0.136 V) being higher than $E^0_{Cd}$ (-0.403 V). Cell formula: $\mathrm{Cd(s)\,\vert\,Cd^{2+}(aq)\,\Vert\,Sn^{2+}(aq)\,\vert\,Sn(s)}$.

Step 2 (standard cell potential). $E^0_{cell} = E^0_{Sn} - E^0_{Cd} = -0.136\ \mathrm{V} - (-0.403\ \mathrm{V}) = 0.267$ V, with $n$ = 2 mol e$^-$.

Step 3 (Gibbs energy). $\Delta G^0 = -nFE^0_{cell} = -2 \times 96500\ \mathrm{C} \times 0.267\ \mathrm{V} = -51531\ \mathrm{V\,C} = -51531\ \mathrm{J} = -51.53$ kJ. …

Identify anode/cathode from the given reaction to get E0cell = E0cathode - E0anode and n = 2, then apply DeltaG0 = -nFE0cell and log10 K = n …

• Computing E0cell = -0.403 - (-0.136) = -0.267 V by treating Cd as the cathode -- the GIVEN reaction oxidises Cd.
• Quoting DeltaG0 in J but labelling it kJ (or vice versa): -51531 J = -51.53 kJ. …