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

$$\begin{gathered} {\mathbf{\text{(a)Cr(s) and Cu}}}^{\mathbf{\text{2 +}}}\mathbf{\text{(aq)}} \\ {\mathbf{\text{(b)Sn(s) and Pb}}}^{\mathbf{\text{2 +}}}\mathbf{\text{(aq)}} \end{gathered}$$

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$\text{Ni}\left(\text{s}\right) \text{and} {\text{Ag}}^{\text{+}}\left(\text{aq}\right)$is given below.

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

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

$$\begin{gathered} \text{2Cr}\left(\text{s}\right)\to {\text{2Cr}}^{\text{+ 3}}\left(\text{aq}\right){\text{+ 6e}}^{\text{-}} {\text{E}°}_{\text{anode}}=-0.73 \text{V} \\ {\text{3Cu}}^{\text{+ 2}}\left(\text{aq}\right){\text{+ 6e}}^{\text{-}}\to 3\text{Cu}\left(\text{s}\right) {\text{E}°}_{\text{cathode}}=0.34 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=0.34V-\left(-0.73V\right) \\ {E^{\circ }}_{cell}=1.07V \end{gathered}$$

We know that.

$$\begin{gathered} △G^{\circ }=-nF{E^{\circ }}_{cell} \\ △G^{\circ }=-6\times 96500C/mol\times 1.07V \\ △G^{\circ }=-619.53KJ/mol \end{gathered}$$

Hence$△G^{\circ }=-619.53KJ/mol$.

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

$Sn\left(s\right)+P{b}^{2+}\left(aq\right)\to S{n}^{2+}\left(aq\right)+Pb\left(s\right)$

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

$$\begin{gathered} \text{Sn}\left(\text{s}\right)\to {\text{Sn}}^{\text{+ 2}}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}} {\text{E}°}_{\text{anode}}\text{=} \text{- 0.14} \text{V} \\ {\text{Pb}}^{\text{+ 2}}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}}\to \text{Pb}\left(\text{s}\right) {\text{E}°}_{\text{cathode}}=-0.13 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=-0.13V-\left(-0.14V\right) \\ {E^{\circ }}_{cell}=0.01V \\ △G^{\circ }=-2\times 96500C/mol\times 0.01V \\ △G^{\circ }=-1.93KJ/mol \end{gathered}$$

Hence$△G^{\circ }=-1.93KJ/mol$.