Question

Calculate standard Gibbs energy change at 25°C for the cell reaction \( \text{Cd(s)} + \text{Sn}^{2+}(\text{aq}) \to \text{Cd}^{2+}(\text{aq}) + \text{Sn(s)} \), \( E^\circ_{\text{Cd}} = -0.403 \, \text{V} \), \( E^\circ_{\text{Sn}} = -0.136 \, \text{V} \).

Hint:

Use \( \Delta G^\circ = -n F E^\circ \) for electrochemical cells; positive \( E^\circ \) indicates spontaneity (\( \Delta G^\circ<0 \)).

Answer 1

Cell: \( \text{Cd} | \text{Cd}^{2+} || \text{Sn}^{2+} | \text{Sn} \).
Anode: \( \text{Cd} \to \text{Cd}^{2+} + 2e^- \), cathode: \( \text{Sn}^{2+} + 2e^- \to \text{Sn} \).
\( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.136 - (-0.403) = 0.267 \, \text{V} \).
Number of electrons \( n = 2 \).
Gibbs energy: \( \Delta G^\circ = -n F E^\circ_{\text{cell}} \), \( F = 96485 \, \text{C mol}^{-1} \), T = 25°C = 298 K (not needed for standard conditions).
\[ \Delta G^\circ = -2 \times 96485 \times 0.267 \approx -51529.59 \, \text{J mol}^{-1} = -51.53 \, \text{kJ mol}^{-1}. \] % Quick tip