Question

Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contained $0.1\,{\text{M}}\,\,{\text{MnO}}_4^ -$ and ${\text{1}}{\text{.72}}\,{\text{M}}\,\,{{\text{H}}^{\text{ + }}}$, and was treated with ${\text{F}}{{\text{e}}^{{\text{2 + }}}}$, necessary to reduce 90% of ${\text{Mn}}{{\text{O}}_{\text{4}}}$ to ${\text{M}}{{\text{n}}^{2 + }}$. $\left[ {E_{Mn{O_4} - /Mn}^0 = \,1.15\,V} \right]$
A. ${\text{1}}{\text{.4V}}$
B. ${\text{1}}{\text{.5V}}$
C. ${\text{1}}{\text{.6V}}$
D. ${\text{1}}{\text{.3V}}$

Answer 1

The half-cell on which a hydrogen electrode is based is given as follows,
$2{{\text{H}}^{\text{ + }}}\left( {aq} \right) + 2{{\text{e}}^ - }\, \to {{\text{H}}_{\text{2}}}\left( g \right)$
In a hydrogen electrode, the electrode is dipped in a solution of acid and hydrogen gas is then bubbled through the electrode. The concentration of both the oxidised and that of the reduced form is unity. .

Complete step by step answer:
We know that Nernst equation is given as,
$E = {E^0} - \dfrac{{0.0592\,V}}{n}{\log _{10}}Q$ ……(1)
Here,
$Q$= reaction quotient
$n$=
${E^0}$= electrode potential
Given that,

$$E_{Mn{O_4} - /Mn}^0 = \,1.15\,V \\\ \Rightarrow \left[ {M{n^{2 + }}} \right] = \,0.09\,V \\\ \Rightarrow \left[ {MnO_4^ - } \right] = \,0.01V \\\ \left[ {{H^ + }} \right] = \,0.08\,V \\\$$

Using this value in equation (1) we will calculate the value of $E$ after 90 % completion as follows.

$${\text{E}} = {{\text{E}}^0} - \dfrac{{{\text{0}}{\text{.059}}}}{{\text{n}}}{\text{log}}\dfrac{{\left[ {{\text{M}}{{\text{n}}^{{\text{2 + }}}}} \right]}}{{\left[ {{\text{MnO}}_{\text{4}}^{\text{ - }}} \right]{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}^{\text{8}}}}} \\\ {\text{E}} = 1.51 - \dfrac{{{\text{0}}{\text{.059}}}}{{\text{n}}}{\text{log}}\dfrac{{0.09}}{{0.01\, \times \,{{0.08}^8}\,}} \\\ \Rightarrow {\text{E}} = 1.51 - \dfrac{{0.059}}{5}\left( {\log \,9 - \,8\,\log \,0.08} \right) \\\ \Rightarrow {\text{E}} = \,1.51 - \dfrac{{0.059}}{5}\left( {9.73} \right) \\\ {\text{E}} = \left( {1.51 - 0.115} \right) = \,1.395\,{\text{V}} \\\$$

As we see that the required value of the potential of an indicator electrode versus the standard hydrogen electrode is $1.395\,{\text{V}} \approx {\text{1}}{\text{.4}}\,{\text{V}}$,

So, Option A is correct.

Additional Information:
We know that the Standard Hydrogen electrode is written in short as SHE. The standard electrode potential is 0 when the temperature is 298 K. SHE is treated as the reference electrode. It helps in comparing the other existing electrodes.
The following reaction take place at the half cell of the SHE
$2{{\text{H}}^{\text{ + }}}\left( {aq} \right) + 2{{\text{e}}^ - }\, \to {{\text{H}}_{\text{2}}}\left( g \right)$
This reaction takes place on a platinum electrode. This half-cell has 1 bar pressure.
For the construction of a standard hydrogen electrode, a platinum electrode, acid solution that have 1 mol per cubic decimetre of ${{\text{H}}^{\text{ + }}}$ molarity, a hydrosol, salt bridge etc. are required.

Note: The platinum electrode is covered with powdered platinum black. The half-cell of the Galvanic cell is attached to the Standard Hydrogen. It helps in making a conductive path ionically through a reservoir. This can also be done by using a salt bridge.