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

Hint: 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.