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.

$△G^{∘}=-nF{E^{∘}}_{cell}$

Where,

n = number of electrons involved in the redox reaction.

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

Number of electrons $\left(\text{n}\right)=1$.

Equilibrium constant at standard conditions$\left(\text{K}\right)=5.04×{10}^4$.

We know that,

$$\begin{gathered} {\text{E}}_{\text{cell}}°=  \frac{\text{RT}}{\text{nF}}\text{lnK} \\ {\text{E}}_{\text{cell}}°=\frac{8.314J/Kmol×298K}{n×96500C/mol}\ln K \\ {\text{E}}_{\text{cell}}°=\frac{0.0592V}{1}\ln K \end{gathered}$$

Putting values of n and K, we get the value of ${\text{E}}_{\text{cell}}°$.

_{$$\begin{gathered} {\text{E}}_{\text{cell}}°=0.0592V×\log 5.0×{10}^{14} \\ {\text{E}}_{\text{cell}}°=0.0592V×4.699 \\ {\text{E}}_{\text{cell}}°=0.278V \end{gathered}$$}

Also,

$△G^{∘}=-nF{\text{E}}_{\text{cell}}°$

Here, _{$n=1,{\text{E}}_{\text{cell}}°=0.278V,F=96500C/mol$}

Therefore,

_{$$\begin{gathered} △G^{∘}=-1×96500C/mol×0.278V \\ △G^{∘}=-26.827KJ/mol \end{gathered}$$.}