91Ó°ÊÓ

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$△G^{∘}$ and the standard electrode potential is given below.

$△G^{∘}=-nF{E^{∘}}_{cell}$

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.

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

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

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

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

We know that.

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

Hence$△G^{∘}=-619.53KJ/mol$.

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

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

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

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

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

Hence$△G^{∘}=-1.93KJ/mol$.