Question

Calculate emf of the following cell at ${\text{25}}{\,^{\text{o}}}{\text{C}}$
${\text{Fe}}/{\text{F}}{{\text{e}}^{{\text{2 + }}}}\,\left( {0.001\,{\text{M}}} \right){\text{/}}{{\text{H}}^ + }\left( {0.01\,{\text{M}}} \right){\text{/}}{{\text{H}}_2}\left( {\text{g}} \right)\left( {{\text{1}}\,{\text{bar}}} \right){\text{/Pt}}\left( {\text{s}} \right)\,\,$
${\text{F}}{{\text{e}}^{{\text{2 + }}}}/{\text{Fe}}\, = \, - 0.44\,{\text{V}}$,${{\text{H}}^ + }/{{\text{H}}_2}\, = \,0.00\,{\text{V}}$

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.
Standard reduction potential ${\text{E}}_{{\text{cell}}}^ \circ \,$is calculated by subtracting the reduction potential of anode from reduction potential of cathode.
${\text{E}}_{{\text{cell}}}^ \circ \, = \,{{\text{E}}_{\text{c}}} - {{\text{E}}_{\text{a}}}$
Substitute $0.00\,{\text{V}}$for ${{\text{E}}_{\text{c}}}$and $- 0.44\,{\text{V}}$for${{\text{E}}_{\text{a}}}$.
${\text{E}}_{{\text{cell}}}^ \circ \, = \,0.00\,{\text{V}} - \left( { - 0.44\,{\text{V}}} \right)$
${\text{E}}_{{\text{cell}}}^ \circ \, = \,0.44\,{\text{V}}$
We can write the half-cell reaction to determine the oxidised and reduced species.
The half-cell reaction are as follows:
Oxidation reaction:${\text{Fe}}\left( {\text{s}} \right)\, \to {\text{F}}{{\text{e}}^{2 + }}\left( {0.001\,{\text{M}}} \right){\text{ + }}\,\,{\text{2}}{{\text{e}}^ - }\,$
Iron is releasing two electrons to form iron ions.
Reduction reaction: ${\text{2}}{{\text{H}}^{\text{ + }}}\left( {0.01\,{\text{M}}} \right)\,{\text{ + }}\,\,{\text{2}}{{\text{e}}^ - }\, \to {{\text{H}}_2}\left( {\text{g}} \right)$
Hydrogen ions are accepting two electrons to form hydrogen gas.
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:
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^ \circ \, - \dfrac{{0.0591}}{{\text{n}}}\log \,\dfrac{{\left[ {{\text{F}}{{\text{e}}^{2 + }}} \right]}}{{{{\left[ {{{\text{H}}^ + }} \right]}^2}}}$
Substitute $0.44\,{\text{V}}$ for${\text{E}}_{{\text{cell}}}^ \circ \,$, $2$for number of electrons, $0.001\,{\text{M}}$ for the concentration of iron ion and $0.01\,{\text{M}}$for the concentration of hydrogen ion.
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,0.44\,{\text{V}}\, - \dfrac{{0.0591}}{2}\log \,\dfrac{{\left[ {0.001} \right]}}{{{{\left[ {0.01} \right]}^2}}}$
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,0.44\,{\text{V}} - 0.030$
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,0.41\,{\text{V}}$

Therefore, the emf of the cell is $0.41\,{\text{V}}$.

Note: The stoichiometry does not affect standard reduction potential. The species which get high reduction potential will reduce and the species which has low reduction potential will get oxidised. In the cell representation, the left side electrode represents the anode and right side represents the cathode.