What are ${{\mathbf{E}}^{\mathbf{\circ }}}_{\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{l}}$and $\mathbf{△}{\mathit{G}}^{\mathbf{\circ }}$of a redox reaction at ${\mathbf{25}}^{\mathbf{0}}\mathit{C}$for which $\mathit{n}\mathbf{=}\mathbf{1}$and$\mathit{K}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$?

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

$△G^{\circ }=-nF{E^{\circ }}_{cell}$

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 $\left(\text{K}\right)=5.04\times {10}^6$.

We know that,

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

${E^{\circ }}_{cell}=\frac{\text{0.0592} \text{V}}{\text{n}}\text{logK}$

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

$$\begin{gathered} {\text{E}}_{\text{cell}}^{\text{o}}\text{=}\frac{0.0592V}{1}{\text{log5.0×10}}^6 \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{=0.0592V×-5.3} \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{=-0.314V} \end{gathered}$$

Also,

$△G^{\circ }=-nF{E^{\circ }}_{cell}$

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

Therefore,

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