Question

Calculate the emf of the cell in which of the following reaction:
${\text{Ni}}\left( {\text{s}} \right)\,{\text{ + }}\,\,{\text{2A}}{{\text{g}}^{\text{ + }}}\left( {0.002\,{\text{M}}} \right)\, \to {\text{N}}{{\text{i}}^{2 + }}\left( {0.160\,{\text{M}}} \right){\text{ + }}\,\,{\text{2Ag}}\,\left( {\text{s}} \right)\,\,$
(Given that ${\text{E}}_{{\text{cell}}}^ \circ \, = \,1.05\,{\text{V}}$)
A. $0.3\,{\text{V}}$
B. ${\text{0}}{\text{.91}}\,{\text{V}}$
C. ${\text{0}}{\text{.62}}\,{\text{V}}$
D. ${\text{0}}{\text{.34}}\,{\text{V}}$

Answer 1

We can calculate the emf of the cell by using Nernst equation. Nernst equation tells the relation in electromotive force (emf) and concentration of oxidised and reduced species.

Formula used: ${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^ \circ \, - \dfrac{{0.0591}}{{\text{n}}}\log \,\dfrac{{\left[ {{\text{oxidized}}} \right]}}{{\left[ {{\text{reduced}}} \right]}}$

Complete step-by-step answer:
The Nernst equation is used to determine the electromotive force (emf) of a cell. The Nernst is represented as follows:
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^ \circ \, - \dfrac{{0.0591}}{{\text{n}}}\log \,\dfrac{{\left[ {{\text{reduced}}} \right]}}{{\left[ {{\text{oxidized}}} \right]}}$
${{\text{E}}_{{\text{cell}}\,}}$ is the electromotive force of the cell.
${\text{E}}_{{\text{cell}}}^ \circ \,$is the standard reduction potential of the cell.
${\text{n}}$is the number of electrons involved in a redox reaction.

We can write the half-cell reaction to determine the oxidised and reduced species.
The half-cell reaction are as follows:
Oxidation reaction:${\text{Ni}}\left( {\text{s}} \right)\, \to {\text{N}}{{\text{i}}^{2 + }}\left( {0.160\,{\text{M}}} \right){\text{ + }}\,\,{\text{2}}{{\text{e}}^ - }\,$
Nickel is releasing two electrons to form nickel ions.
Reduction reaction: ${\text{2A}}{{\text{g}}^{\text{ + }}}\left( {0.002\,{\text{M}}} \right)\,{\text{ + }}\,\,{\text{2}}{{\text{e}}^ - }\, \to {\text{2Ag}}\,\left( {\text{s}} \right)$
Silver ions are accepting two electrons to form silver metal.
So, the number of electrons exchanged during the reaction is$2$.
So, the Nernst equation for the nickel and silver redox reaction is as follows:
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^ \circ \, - \dfrac{{0.0591}}{{\text{n}}}\log \,\dfrac{{\left[ {{\text{N}}{{\text{i}}^{2 + }}} \right]}}{{{{\left[ {{\text{A}}{{\text{g}}^ + }} \right]}^2}}}$

On substituting $1.05\,{\text{V}}$ for${\text{E}}_{{\text{cell}}}^ \circ \,$, $2$for number of electrons, $0.160\,{\text{M}}$ for the concentration of nickel ion and $0.002\,{\text{M}}$for the concentration of silver ion.
$\Rightarrow {{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,1.05\,{\text{V}}\, - \dfrac{{0.0591}}{2}\log \,\dfrac{{\left[ {0.160} \right]}}{{{{\left[ {0.002} \right]}^2}}}$
$\Rightarrow {{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,1.05\,{\text{V}}\, - \dfrac{{0.0591}}{2}\log \,\dfrac{{\left[ {0.160} \right]}}{{\left[ {4 \times {{10}^{ - 6}}} \right]}}$
$\Rightarrow {{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,1.05\,{\text{V}}\, - \dfrac{{0.0591}}{2}\log \,40000$
$\Rightarrow {{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,1.05\,{\text{V}}\, - \dfrac{{0.0591}}{2} \times 4.60$
$\Rightarrow {{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,1.05\,{\text{V}} - 0.14$
$\Rightarrow {{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,0.91\,{\text{V}}$
So, the emf of the cell is $0.91\,{\text{V}}$.

Therefore, option (B) $0.91\,{\text{V}}$, is correct.

Note: The stoichiometry does not affect standard reduction potential. Standard reduction potential ${\text{E}}_{{\text{cell}}}^ \circ \,$is calculated by subtracting the reduction potential of anode from reduction potential of cathode.