Q21.67P - Chemistry: Molecular Nature Of Matter And Change

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 $\mathbf{\text{K = 0.065}}$?

Step-by-Step Solution

Answer

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

$△G^{\circ }=6.75 KJ/mol$

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 ∘ c e l l

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

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

We know that,

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

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

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

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

Also,

$△G^{\circ }=-nF{E^{\circ }}_{cell}$

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

Therefore,

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