Calculate $\mathbf{△}{\mathit{G}}^{\mathbf{\circ }}$for each of the reactions

$$\begin{gathered} {\mathbf{\text{(a)Al(s) and Cd}}}^{\mathbf{\text{2 +}}}\mathbf{\text{(aq)}} \\ {\mathbf{\text{(b)I}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s) and Br}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq)}} \end{gathered}$$

For a redox reaction taking place, the standard reduction potential is the difference between the respective cell potential.

${\text{E}}_{\text{cell}}°={\text{E}}_{\text{cathode}}°{\text{- E}}_{\text{anode}}°$

The relation between $△G^{\circ }$ and the standard electrode potential is given below.

$△G^{\circ }=-nF{E^{\circ }}_{cell}$

Where,

n = number of electrons involved in the redox reaction.

$\text{F} \text{=} \text{96500} \text{C/mol}$.

The redox reaction taking place between $\text{Al}\left(\text{s}\right) \text{and} {\text{Cd}}^{\text{+ 2}}\left(\text{aq}\right)$ is given below.

$2Al\left(\text{s}\right)+3{\text{Cd}}^{2+}\text{(aq)}\to 2{\text{Al}}^{3+}\left(\text{aq}\right)+3\text{Cd}\left(\text{s}\right)$

Now, we have to write down the cathode and anode half-cell reaction along with their respective electrode potential.

$$\begin{gathered} \text{2Al}\left(\text{s}\right)\to 2{\text{Al}}^{\text{+ 3}}\left(\text{aq}\right){\text{+ 6e}}^{\text{-}} {\text{E}°}_{\text{anode}}=-1.66 V \\ {\text{3Cd}}^{\text{+ 2}}\left(\text{aq}\right){\text{+ 6e}}^{\text{-}}\to 2\text{Cd}\left(\text{s}\right) {{\text{E}}^{\circ \pm }}_{\text{cathode}}=-0.40 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=-0.40V-\left(-1.66\right)V \\ {E^{\circ }}_{cell}=1.26V \end{gathered}$$

We know that,

$$\begin{gathered} △G^{\circ }=-nF{E^{\circ }}_{cell} \\ △G^{\circ }=-6\times 96500C/mol\times 1.26V \\ △G^{\circ }=-729.54KJ/mol \end{gathered}$$

Hence$△G^{\circ }=-729.54KJ/mol$

The redox reaction taking place between ${\text{l}}_2\left(\text{s}\right) \text{and} {\text{Br}}^{\text{-}}\left(\text{aq}\right)$is given below.

${\text{I}}_2(\text{s})+2{\text{Br}}^{-}\left(\text{aq}\right)\to 2{\text{I}}^{-}\left(\text{aq}\right)+{\text{Br}}_2\left(\text{l}\right)$

Now, we have to write down the cathode and anode half-cell reaction along with their respective electrode potential.

$$\begin{gathered} {\text{l}}_2\left(\text{s}\right){\text{+ 2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}\left(\text{aq}\right) {\text{E}°}_{\text{cathode}}=0.53\text{V} \\ {\text{2Br}}^{\text{-}}\left(\text{aq}\right) \to {\text{Br}}_{\text{2}}\left(\text{s}\right){\text{+ 2e}}^{\text{-}} {\text{E}°}_{\text{anode}}=1.07 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=\left(0.53-1.07\right)V \\ {E^{\circ }}_{cell}=-0.54V \end{gathered}$$

We know that,

$$\begin{gathered} △G^{\circ }=-nF{E^{\circ }}_{cell} \\ △G^{\circ }=-2\times 96500C/mol\times -0.54V \\ △G^{\circ }=104.22KJ/mol \end{gathered}$$

Hence $△G^{\circ }=104.22KJ/mol$.