Question

What are ${{\mathit{E}}^{\mathbf{\circ }}}_{\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{l}}$ 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$\mathit{K}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}$.

Short answer

${E^0}_{cell}=0.278V$

$△G^{\circ }=-26.827KJ/mol$

Step-by-step solution

Standard electrode potential, equilibrium constant (K) and △G∘.

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.

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

Calculation of  △E∘cell.

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

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

We know that,

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

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

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

Also,

$△G^{\circ }=-nF{\text{E}}_{\text{cell}}°$

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

Therefore,

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