What is standard decrease in Gibbs energy for the following cell?...

The standard cell potential is given by:

E^{}^{\circ}_{\text{cell}} = E^{}^{\circ}_{\text{cathode}} - E^{}^{\circ}_{\text{anode}}

E^{}^{\circ}_{\text{cell}} = -0.40 - (-1.66) = 1.26 \text{ V}

The half-cell reactions are:

Anode: $\text{Al} \rightarrow \text{Al}^{+3} + 3e^-$

Cathode: $\text{Cd}^{+2} + 2e^- \rightarrow \text{Cd}$

To balance the number of electrons, the overall reaction involves the transfer of $6$ electrons, so $n = 6$.

The standard Gibbs free energy change is:

\Delta G^{}^{\circ} = -n F E^{}^{\circ}_{\text{cell}}

\Delta G^{}^{\circ} = -6 \times 96500 \times 1.26

\Delta G^{}^{\circ} = -729540 \text{ J} = -729.54 \text{ kJ}

Answer: $-729.54$ kJ