91影视

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

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

The relation between equilibrium constant and the standard electrode potential is given below.

${\text{E}}_{\text{cell}}掳=鈥�鈥�\frac{\text{RT}}{\text{nF}}\text{lnK}$

The relation between and the standard electrode potential is given below.

$鈻�G^{鈭�}=-nF{E^{鈭�}}_{cell}$

Where,

n = number of electrons involved in the redox reaction

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

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

Equilibrium constant at standard conditions $\left(\text{K}\right)=65$.

We know that,

$$\begin{gathered} {\text{E}}_{\text{cell}}掳=鈥�鈥�\frac{\text{RT}}{\text{nF}}\text{lnK} \\ {\text{E}}_{\text{cell}}掳=\frac{8.314J/Kmol脳298K}{n脳96500C/mol}\log K \end{gathered}$$

${\text{E}}_{\text{cell}}掳=\frac{0.0592V}{n}\log K$

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

$$\begin{gathered} {\text{E}}_{\text{cell}}掳=\frac{0.0592V}{2}\log 65 \\ {\text{E}}_{\text{cell}}掳=\frac{0.0592V}{2}脳1.813 \\ {\text{E}}_{\text{cell}}掳=0.054V \end{gathered}$$

Also,

$鈻�G^{鈭�}=-nF{E^{鈭�}}_{cell}$

Here,$n=2,{E^{鈭�}}_{cell}=0.054V,andF=96500C/mol$.

Therefore,

$$\begin{gathered} 鈻�G^{鈭�}=-2脳96500脳0.054 \\ 鈻�G^{鈭�}=-10.42KJ/mol \end{gathered}$$.