For the galvanic cell reaction given below: 2Cr(s) + 3Cd^ 2+ (aq)...

The given cell reaction is:

$2\mathrm{Cr(s)} + 3\mathrm{Cd}^{2+}\mathrm{(aq)} \rightarrow 2\mathrm{Cr}^{3+}\mathrm{(aq)} + 3\mathrm{Cd(s)}$

The number of electrons transferred in the balanced reaction is $n = 6$.

The standard Gibbs free energy change is given by the relation:

$\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$

Substituting the given values:

$\Delta G^{\circ} = -6 \times 96500 \times 0.34$

$\Delta G^{\circ} = -196860\text{ J mol}^{-1}$

$\Delta G^{\circ} = -196.86\text{ kJ mol}^{-1}$

Answer: $-196.86\text{ kJ mol}^{-1}$