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 equilibrium constant and the standard electrode potential is given below.

${\text{E}}_{\text{cell}}°=  \frac{\text{RT}}{\text{nF}}\text{lnK}$

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

$\text{△G}°={\text{- nFE}}_{\text{cell}}°$

Where,

n = number of electrons involved in the redox reaction

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

Number of electrons $\left(\text{n}\right)=2$.

Equilibrium constant at standard conditions $\left(\text{K}\right)=0.065$.

We know that,

$$\begin{gathered} {\text{E}}_{\text{cell}}°=  \frac{\text{RT}}{\text{nF}}\text{lnK} \\ {\text{E}}_{\text{cell}}°=\frac{8.314J/Kmol×298K}{n×96500C/mol}\ln K \end{gathered}$$

${E^{∘}}_{cell}=\frac{0.0592V}{n}\log K$

Putting values of n and K, we get the value of ${E^{∘}}_{cell}$.

$$\begin{gathered} {E^{∘}}_{cell}=\frac{0.0592V}{2}\log 0.065 \\ {E^{∘}}_{cell}=\frac{0.0592V}{2}×-1.18 \\ {E^{∘}}_{cell}=-0.035V \end{gathered}$$

Also,

$△G^{∘}=-nF{E^{∘}}_{cell}$

$Here,n=2,{E^{∘}}_{cell}=-0.035VandF=96500C/mol$.

Therefore,

$$\begin{gathered} △G^{∘}=-2×96500×-0.035 \\ △G^{∘}=6.75KJ/mol \end{gathered}$$.