Question

The emf of the cell ${\text{F}}{{\text{e}}^{\text{ + }}}\,{\text{ + }}\,\,{\text{Zn}}\, \to {\text{Z}}{{\text{n}}^{2 + }}{\text{ + }}\,{\text{Fe}}\,$is:
A. $+ 0.35\,{\text{volt}}$
B. $- 0.35\,{\text{volt}}$
C. $+ \,{\text{1}}{\text{.17}}\,{\text{volt}}$
D. $- \,{\text{1}}{\text{.17}}\,{\text{volt}}$

Answer 1

Standard reduction potential of the half-cell is given and we have to determine the standard reduction potential of the cell. So, we should know the relation between the standard reduction potential of the half-cell and cell. The standard reduction potential of the cell is determined by subtracting the reduction potential of the anode from the reduction potential of the cathode.

Formula used: ${\text{E}}_{{\text{cell}}}^ \circ {\text{ = }}\,{\text{E}}_{{\text{cathode}}}^ \circ \, - {\text{E}}_{{\text{anode}}}^ \circ$

Complete solution:
The relation between the standard reduction potential of the cell and half-cell is as follows:
${\text{E}}_{{\text{cell}}}^ \circ {\text{ = }}\,{\text{E}}_{{\text{cathode}}}^ \circ \, - {\text{E}}_{{\text{anode}}}^ \circ$
Where,
${\text{E}}_{{\text{cell}}}^ \circ$ is the standard reduction potential of the cell
${\text{E}}_{{\text{cathode}}}^ \circ$is the reduction potential of the cathode half-cell
${\text{E}}_{{\text{anode}}}^ \circ$is the reduction potential of the anode half-cell
The given half-cell reaction and oxidation potential are as follows:
${\text{Zn}}\, \to {\text{Z}}{{\text{n}}^{2 + }} + \,{\text{2}}{{\text{e}}^ - }\,$; ${{\text{E}}^{\text{o}}}\, = \, + 0.7{\text{6}}\,{\text{volt}}$
${\text{Fe}}\, \to {\text{F}}{{\text{e}}^{2 + }} + \,{\text{2}}{{\text{e}}^ - }\,$; ${{\text{E}}^{\text{o}}}\, = \, + 0.41\,{\text{volt}}$
As we compare the reduction potential so, we have to determine the reduction potential values for the given half-cell reaction so, we will write the reduction reaction for both half-cell as follows:
${\text{Z}}{{\text{n}}^{2 + }} + \,{\text{2}}{{\text{e}}^ - }\, \to \,{\text{Zn}}\,$; ${{\text{E}}^{\text{o}}}\, = \, - 0.7{\text{6}}\,{\text{volt}}$
${\text{F}}{{\text{e}}^{2 + }} + \,{\text{2}}{{\text{e}}^ - }\, \to {\text{Fe}}\,$; ${{\text{E}}^{\text{o}}}\, = \, - 0.41\,{\text{volt}}$
Now, we will decide the cathode and anode as follows:
The species having high reduction potential get reduced at the cathode and the species having low reduction potential get oxidized at the anode.
The reduction potential of Fe ($- 0.41\,{\text{volt}}$) is greater than the reduction potential of Zn ($- 0.76\,{\text{volt}}$) so, Fe will get reduce at the cathode and Zn will get oxidized at the anode.
Oxidation reaction:${\text{Zn}}\, \to {\text{Z}}{{\text{n}}^{2 + }}{\text{ + }}\,\,{\text{2}}{{\text{e}}^ - }\,$
Reduction reaction: ${\text{F}}{{\text{e}}^{2 + }}{\text{ + }}\,\,{\text{2}}{{\text{e}}^ - }\, \to {\text{Fe}}\,$
On substituting $- 0.41\,{\text{volt}}$ for${\text{E}}_{{\text{cathode}}}^ \circ$ and$- 0.76\,{\text{volt}}$ for ${\text{E}}_{{\text{anode}}}^ \circ$,
${\text{E}}_{{\text{cell}}}^ \circ {\text{ = }}\, - 0.41\,{\text{volt}}\, - \left( { - 0.76\,{\text{volt}}} \right)$
${\text{E}}_{{\text{cell}}}^ \circ {\text{ = }}\, + 0.35\,{\text{volt}}$
So, the emf of the cell is $+ 0.35\,{\text{volt}}$.

Therefore, the correct answer is (A).

Note: The reduction takes place at the cathode and oxidation takes place at the anode. The stoichiometry does not affect standard reduction potential. On reversing the reaction the sign of standard reduction potential also reverses. On multiplying or dividing the reaction with any coefficient, the standard reduction potential does not get multiplied.