Question

The equilibrium constant for the following reaction:
$H{g^{2 + }} + Hg \to H{g_2}^{2 + }$
${E^ \circ }_{({\text{Hg | H}}{{\text{g}}^{{\text{2 + }}}})} = - 0.788V;$ ${E^ \circ }_{({\text{H}}{{\text{g}}_2}^{2 + }{\text{ | 2H}}{{\text{g}}^{{\text{2 + }}}})} = - 0.92V$ is ${K_{eq}} = 2.98 \times {10^x}$
The value of $x$ is _____.

Answer 1

The relation between ${E^ \circ }_{cell}$ and equilibrium constant i.e. ${K_{eq}}$. The relation is as:
${E^ \circ } = \dfrac{{0.059}}{n}\log {K_{eq}}$ where ${E^ \circ }_{cell}$ is cell voltage, ${K_{eq}}$ is equilibrium constant and $n$ is total number of electrons transfer in the reaction.

Complete step by step solution:
First of all let us talk about how to find ${E^ \circ }_{cell}$ of the reaction when you know the values of ${E^ \circ }$ of its components i.e. the ions or atoms which are reduced or oxidised in the reaction.
Reduced ions or atoms are that in which there is decrease in the positive charge on the ion i.e. gain of electrons.
Oxidised ions or atoms are that in which there is an increase in the positive charge on the ion i.e. loss of electrons.
Reaction given is as:
$H{g^{2 + }} + Hg \to H{g_2}^{2 + }$
Therefore the reaction at anode will be: $Hg \to H{g^{2 + }} + 2{e^ - }$ and also we know the value of the ${E^{^ \circ }}$ as ${E^ \circ }_{({\text{Hg | H}}{{\text{g}}^{{\text{2 + }}}})} = - 0.788V$.
The reaction at cathode will be as: $2H{g^{2 + }} + 2{e^ - } \to H{g_2}^{2 + }$ and also we know the value of ${E^ \circ }$ as ${E^ \circ }_{({\text{H}}{{\text{g}}_2}^{2 + }{\text{ | 2H}}{{\text{g}}^{{\text{2 + }}}})} = - 0.92V$.
Now if in the reaction we know which is oxidised and which is reduced atom and we also know the values of ${E^ \circ }$ of its components i.e. the ions or atoms. Then we can calculate the value of ${E^ \circ }_{cell}$ of the reaction as: $\;{E^ \circ }_{cell} = \;{E^ \circ }_R - \;{E^ \circ }_L$ where ${E^ \circ }_R$ is the ${E^ \circ }$of the atom which is at right side (cathode) and ${E^ \circ }_L$ is the ${E^ \circ }$ of the atom which is at left side (anode). Hence ${E^ \circ } = - 0.788 - ( - 0.92) = 0.132V$.
Now we know the relation between ${E^ \circ }_{cell}$ and equilibrium constant i.e. ${K_{eq}}$. The relation is as:
${E^ \circ } = \dfrac{{0.059}}{n}\log {K_{eq}}$ where ${E^ \circ }_{cell}$ is cell voltage, ${K_{eq}}$ is equilibrium constant and $n$ is total number of electrons transfer in the reaction. In this reaction the total number of electrons transferred are $2$. So $n = 2$. Now we know the value of ${E^ \circ }_{cell}$ so we can calculate the value of ${K_{eq}}$ as:
$\log {K_{eq}} = \dfrac{{2 \times 0.132}}{{0.059}}$. Hence the value of ${K_{eq}}$will be $2.98 \times {10^4}$.

So the value of $x$ is $4$.

Note:
${E_{cell}}$ of a reaction is defined as electrode potential of the cell and ${E^ \circ }_{cell}$ of a reaction is defined as electrode potential measured at $1$ atmosphere pressure, $1$ molar solution at ${25^ \circ }C$ also known as standard electrode potential.