Question

Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contained 0.1 M $Mn{O_4}^ - $ and 0.8 M ${H^ + }$and which was treated with $F{e^{2 + }}$ necessary to reduce 90% of the $Mn{O_4}^ - $ ​ to $M{n^{2 + }}$.
$Mn{O_4}^ - $​+8${H^ + }$+5e→$M{n^{2 + }}$+${H_2}O$, ${E^0}$=1.51V
A. 1.39V
B. 2.39V
C. 1.59V
D. None of these

Answer 1

Hint: Nernst equation allows the accurate determination of equilibrium constants. We should observe the chemical formula and differentiate the reduction and oxidation to apply in the formula. We have to use the Nernst equation to calculate the potential indicator electrode versus the standard hydrogen electrode.

Complete step by step solution:
From the given data:
${E^0}$=1.51V
${H^ + }$=0.8M
$Mn{O_4}^ - $=0.1M
Solving using Nernst equation:
$\
  E = {E^0} - \dfrac{{kT}}{{nF}}\ln \left[ {\dfrac{{\operatorname{Re} d}}{{Ox}}} \right] \\
\implies E = {E^0} - \dfrac{{0.0591}}{n}\log \left[ {\dfrac{{\operatorname{Re} d}}{{Ox}}} \right] \\
\ $
$\implies E = {E^0} - \dfrac{{0.0591}}{n}\log \left[ {\dfrac{{M{n^{2 + }}}}{{\left( {Mn{O_4}^ - } \right).{{({H^ + })}^8}}}} \right]$
$\
\implies E = 1.51 - \dfrac{{0.0591}}{5}\log \left[ {\dfrac{{0.09}}{{0.01 \times {{(0.08)}^8}}}} \right] \\
\implies E = 1.51 - \dfrac{{0.0591}}{5}\log \left[ {\dfrac{9}{{{{(0.08)}^8}}}} \right] \\
\implies E = 1.51 - \dfrac{{0.0591}}{5}(\log 9 - 8\log 0.8) \\
\ $
$\
\implies E = 1.51 - \dfrac{{0.0591}}{5}(9.73) \\
\implies E = (1.51 - 0.115) \\
\therefore E = 1.395V \\
\ $

So, the correct answer is “Option A”.

Note:
Commonly Nernst equations can be used to measure the cell potential at any interval during the reaction or at conditions of standard-state.
Also students should understand how we need to apply the logarithm for a certain value and get the solution.The potential indicator electrode versus the standard hydrogen electrode.