Q21.66P - Chemistry: Molecular Nature Of Matter And Change

Question

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 ${\mathbf{25}}^{\mathbf{\circ }}\mathit{C}$ for which $\mathit{n}\mathbf{=}\mathbf{2}$ and $\mathit{K}\mathbf{=}\mathbf{65}$?

Step-by-Step Solution

Answer

$$\begin{gathered} {E^{\circ }}_{cell}=0.054 V \\ △G^{\circ }=-10.24KJ/mol \end{gathered}$$.

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 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 ∘ c e l l

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

Equilibrium constant at standard conditions $\left(\text{K}\right)=65$.

We know that,

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

 ${\text{E}}_{\text{cell}}°=\frac{0.0592 V}{n}\log K$

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

$$\begin{gathered} {\text{E}}_{\text{cell}}°=\frac{0.0592 V}{2}\log 65 \\ {\text{E}}_{\text{cell}}°=\frac{0.0592 V}{2}\times 1.813 \\ {\text{E}}_{\text{cell}}°=0.054 V \end{gathered}$$

Also,

$△G^{\circ }=-nF{E^{\circ }}_{cell}$

Here, $n=2,{E^{\circ }}_{cell}=0.054 V, and F=96500C/mol$.

Therefore,

$$\begin{gathered} △G^{\circ }=-2\times 96500\times 0.054 \\ △G^{\circ }=-10.42KJ/mol \end{gathered}$$.