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$\mathit{n}\mathbf{=}\mathbf{2}$ and$\mathbf{\text{K = 0.065}}$?

Short answer

${E^{\circ }}_{cell}=-0.035V$

$△G^{\circ }=6.75KJ/mol$

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.

$\text{△G}°={\text{- nFE}}_{\text{cell}}°$

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

Equilibrium constant at standard conditions $\left(\text{K}\right)=0.065$.

We know that,

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

${E^{\circ }}_{cell}=\frac{0.0592V}{n}\log K$

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

$$\begin{gathered} {E^{\circ }}_{cell}=\frac{0.0592V}{2}\log 0.065 \\ {E^{\circ }}_{cell}=\frac{0.0592V}{2}\times -1.18 \\ {E^{\circ }}_{cell}=-0.035V \end{gathered}$$

Also,

$△G^{\circ }=-nF{E^{\circ }}_{cell}$

$Here,n=2,{E^{\circ }}_{cell}=-0.035VandF=96500C/mol$.

Therefore,

$$\begin{gathered} △G^{\circ }=-2\times 96500\times -0.035 \\ △G^{\circ }=6.75KJ/mol \end{gathered}$$.