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$鈻�G^{鈭�}$ 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 $\left(\text{n}\right)=1$.

Equilibrium constant at standard conditions $\left(\text{K}\right)=5.04脳{10}^6$.

We know that,

$$\begin{gathered} {E^{鈭�}}_{cell}=\frac{RT}{nF}\ln K \\ {E^{鈭�}}_{cell}=\frac{8.314J/Kmol}{n脳96500C/mol}\ln K \end{gathered}$$

${E^{鈭�}}_{cell}=\frac{\text{0.0592}鈥�\text{V}}{\text{n}}\text{logK}$

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

$$\begin{gathered} {\text{E}}_{\text{cell}}^{\text{o}}\text{=}\frac{0.0592V}{1}{\text{濒辞驳5.0脳10}}^6 \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{=0.0592痴脳-5.3} \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{=-0.314V} \end{gathered}$$

Also,

$鈻�G^{鈭�}=-nF{E^{鈭�}}_{cell}$

Here, $n=1$, ${E^{鈭�}}_{cell}=-0.314V$, and ${\text{F = 96500Cmol}}^{\text{- 1}}$.

Therefore,

$$\begin{gathered} 鈻�G^{鈭�}=-1.96500*\left(-0.314\right) \\ 鈻�G^{鈭�}=30.30KJ/mol \end{gathered}$$.