Question

${\text{Pb}}\left( {\text{s}} \right) + {\text{H}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons {\text{PbS}}{{\text{O}}_{\text{4}}}\left( {\text{s}} \right) + {\text{2Hg}}\left( {\text{l}} \right)$ design the cell if both electrolytes are present in their saturated solution state. Given ${\text{E}}_{{\text{Pb/P}}{{\text{b}}^{2 + }}}^ \circ$ and ${\text{E}}_{{\text{Hg/Hg}}_2^{2 + }}^ \circ$ are $0.126{\text{ V}}$ and $0.789{\text{ V}}$ respectively and ${{\text{K}}_{{\text{SP}}}}$ of ${\text{PbS}}{{\text{O}}_{\text{4}}}$ and ${\text{H}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ are $2.43 \times {10^{ - 8}}$ and $1.46 \times {10^{ - 6}}$ respectively. The e.m.f. of given cell reaction to the nearest integer.

Answer 1

We are given a redox reaction in which ${\text{Pb}}$ is getting oxidised and ${\text{Hg}}$ is getting reduced. Calculate the e.m.f. of a given cell using Nernst's equation. Also, we are given the solubility product constant. From the solubility product constant, calculate the concentration of the respective ions.

Formulae Used:
${\text{E}}_{{\text{cell}}}^ \circ = {\text{E}}_{{\text{reduction}}}^ \circ - {\text{E}}_{{\text{oxidation}}}^ \circ$
${{\text{E}}_{{\text{cell}}}} = {\text{E}}_{{\text{cell}}}^ \circ - \dfrac{{{\text{0}}{\text{.059}}}}{{\text{n}}}\log \dfrac{{[{\text{Reduction}}]}}{{[{\text{Oxidation}}]}}$

Complete answer:
We are given the reaction,
${\text{Pb}}\left( {\text{s}} \right) + {\text{H}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons {\text{PbS}}{{\text{O}}_{\text{4}}}\left( {\text{s}} \right) + {\text{2Hg}}\left( {\text{l}} \right)$
We know that the oxidation reaction occurs at the anode and the reduction reaction occurs at the cathode. Thus, the reactions occurring at anode and cathode are,
${\text{Anode: Pb}} \to {\text{P}}{{\text{b}}^{2 + }} + {\text{2}}{{\text{e}}^ - }$
${\text{Cathode: Hg}}_2^{2 + } + {\text{2}}{{\text{e}}^ - } \to 2{\text{Hg}}$

In the reaction, ${\text{Pb}}$ is getting oxidised and ${\text{Hg}}$ is getting reduced. Thus, ${\text{E}}_{{\text{Pb/P}}{{\text{b}}^{2 + }}}^ \circ = 0.126{\text{ V}}$ is the oxidation potential and ${\text{E}}_{{\text{Hg/Hg}}_2^{2 + }}^ \circ = 0.789{\text{ V}}$ is the reduction potential.

The dissociation of ${\text{PbS}}{{\text{O}}_{\text{4}}}$ occurs as follows:
${\text{PbS}}{{\text{O}}_{\text{4}}} \rightleftharpoons {\text{P}}{{\text{b}}^{2 + }} + {\text{SO}}_4^{2 - }$

We know that the solubility product constant is the product of concentration of the individual ions. Thus,
$\Rightarrow {{\text{K}}_{{\text{SP}}}} = [{\text{P}}{{\text{b}}^{2 + }}{\text{][SO}}_4^{2 - }]$
$\Rightarrow {{\text{K}}_{{\text{SP}}}} = {[{\text{P}}{{\text{b}}^{2 + }}{\text{]}}^2}$ …… $[{\text{P}}{{\text{b}}^{2 + }}{\text{]}} = {\text{[SO}}_4^{2 - }]$
We are given that the solubility product constant $\left( {{{\text{K}}_{{\text{SP}}}}} \right)$ for $\Rightarrow {\text{PbS}}{{\text{O}}_{\text{4}}}$ is $2.43 \times {10^{ - 8}}$. Thus,
$\Rightarrow {[{\text{P}}{{\text{b}}^{2 + }}{\text{]}}^2} = 2.43 \times {10^{ - 8}}$
$[{\text{P}}{{\text{b}}^{2 + }}{\text{]}} = \sqrt {2.43 \times {{10}^{ - 8}}}$
The dissociation of ${\text{H}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ occurs as follows:
${\text{H}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons {\text{Hg}}_2^{2 + } + {\text{SO}}_4^{2 - }$
We know that the solubility product constant is the product of concentration of the individual ions. Thus,
$\Rightarrow {{\text{K}}_{{\text{SP}}}} = [{\text{Hg}}_2^{2 + }][{\text{SO}}_4^{2 - }]$
$\Rightarrow {{\text{K}}_{{\text{SP}}}} = {[{\text{Hg}}_2^{2 + }]^2}$ …… $[{\text{Hg}}_2^{2 + }] = [{\text{SO}}_4^{2 - }]$
We are given that the solubility product constant $\left( {{{\text{K}}_{{\text{SP}}}}} \right)$ for ${\text{H}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $1.46 \times {10^{ - 6}}$. Thus,
$\Rightarrow {[{\text{Hg}}_2^{2 + }]^2} = 1.46 \times {10^{ - 6}}$
$\Rightarrow [{\text{Hg}}_2^{2 + }] = \sqrt {1.46 \times {{10}^{ - 6}}}$

We can calculate the standard potential of the cell using the equation as follows:
$\Rightarrow {\text{E}}_{{\text{cell}}}^ \circ = {\text{E}}_{{\text{reduction}}}^ \circ - {\text{E}}_{{\text{oxidation}}}^ \circ$
In the reaction, ${\text{Pb}}$ is getting oxidised and ${\text{Hg}}$ is getting reduced. Thus,
$\Rightarrow {\text{E}}_{{\text{cell}}}^ \circ = {\text{E}}_{{\text{Hg/Hg}}_2^{2 + }}^ \circ - {\text{E}}_{{\text{Pb/P}}{{\text{b}}^{2 + }}}^ \circ$
Substitute $0.126{\text{ V}}$ for ${\text{E}}_{{\text{Pb/P}}{{\text{b}}^{2 + }}}^ \circ$ and $0.789{\text{ V}}$ for ${\text{E}}_{{\text{Hg/Hg}}_2^{2 + }}^ \circ$. Thus,
$\Rightarrow {\text{E}}_{{\text{cell}}}^ \circ = 0.789{\text{ V}} - 0.126{\text{ V}}$
$\Rightarrow {\text{E}}_{{\text{cell}}}^ \circ = 0.663{\text{ V}}$
Thus, the standard potential of the cell is $0.663{\text{ V}}$.
We know the Nernst equation is as follows:
${{\text{E}}_{{\text{cell}}}} = {\text{E}}_{{\text{cell}}}^ \circ - \dfrac{{{\text{0}}{\text{.059}}}}{{\text{n}}}\log \dfrac{{[{\text{Reduction}}]}}{{[{\text{Oxidation}}]}}$
Where ${{\text{E}}_{{\text{cell}}}}$ is the potential of cell,
${\text{E}}_{{\text{cell}}}^ \circ$ is the standard potential of cell,
${\text{n}}$ is the number of moles of an electron.
In the reaction, ${\text{Pb}}$ is getting oxidised and ${\text{Hg}}$ is getting reduced. Thus,
$\Rightarrow {{\text{E}}_{{\text{cell}}}} = {\text{E}}_{{\text{cell}}}^ \circ - \dfrac{{{\text{0}}{\text{.059}}}}{{\text{n}}}\log \dfrac{{[{\text{Hg}}_2^{2 + }]}}{{[{\text{P}}{{\text{b}}^{2 + }}]}}$
Substitute $0.663{\text{ V}}$ for the standard potential of the cell, 2 for the number of moles of electron, $\sqrt {1.46 \times {{10}^{ - 6}}}$ for the $[{\text{Hg}}_2^{2 + }]$ and $\sqrt {2.43 \times {{10}^{ - 8}}}$ for the $[{\text{P}}{{\text{b}}^{2 + }}{\text{]}}$. Thus,
$\Rightarrow {{\text{E}}_{{\text{cell}}}} = 0.663{\text{ V}} - \dfrac{{{\text{0}}{\text{.059}}}}{{\text{2}}}\log \dfrac{{\sqrt {1.46 \times {{10}^{ - 6}}} }}{{\sqrt {2.43 \times {{10}^{ - 8}}} }}$
$\Rightarrow {{\text{E}}_{{\text{cell}}}} = 0.663{\text{ V}} - \dfrac{{{\text{0}}{\text{.059}}}}{{\text{2}}} \times 0.8893$
$\Rightarrow {{\text{E}}_{{\text{cell}}}} = 0.663{\text{ V}} - 0.02623$
$\Rightarrow {{\text{E}}_{{\text{cell}}}} = 0.6367{\text{ V}}$

Thus, the e.m.f. of the cell is $0.6367{\text{ V}}$.

Note: The Nernst equation is applicable to single electrode reduction or oxidation potentials at any condition, standard electrode potentials, electromotive force of an electrochemical cell, unknown ionic concentrations, determining feasibility of electrochemical cells, etc. The equation is not applicable to the solutions having high concentrations and the equation cannot be used to measure the cell potential when current is flowing through the electrode.