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If the ${H^ + }$ concentration decreased from $1M$ to ${10^{ - 4}}M$ at ${25^\circ }C$ for the couple $Mn{O^{ - 4}}$/$M{n^{2 + }}$, then the oxidising power of the $Mn{O^{ - 4}}$/$M{n^{2 + }}$ couple decreases by how much?

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Last updated date: 22nd Feb 2024
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Hint: As we know that in an electrochemical cell, there is a separation of charges between the electrodes and the solution that set up the difference in potential of both and the oxidising power or oxidising potential of an electrode is the tendency to lose electrons or to get oxidised.

Complete step by step solution:
We are well aware with the redox potentials of electrochemical cell where the oxidation potential is defined as the tendency of the electrode to lose electrons or gets oxidised whereas the reduction potential is defined as the tendency of an electrode to simply gain those electrons or to get reduced.
Now we also know that the permanganate reacts with hydrogen ions, releasing five electrons and gives elemental manganese as product on completion of reaction and the half-cell reaction of the given couple is shown as:
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
We also know the formula for calculating the electrode potential which can be given as:
${E_{cell}} = {E^\circ } - \dfrac{{0.0591}}{n}\log {K_c}$, where $n$ is the number of electrons.
We are given the concentration of $Mn{O^{ - 4}}$, using these concentrations we can first identify the change in the electrode potential. So, using the above formula, the initial electrode potential of $Mn{O^{ - 4}}$ when the concentration is $1M$, we will get:
${E_1} = {E^\circ } - \dfrac{{0.0591}}{5}\log \dfrac{{[M{n^{2 + }}]}}{{{{(1)}^8}}}$
Similarly, the final electrode potential will be given as:
${E_2} = {E^\circ } - \dfrac{{0.0591}}{5}\log \dfrac{{[M{n^{2 + }}]}}{{{{({{10}^{ - 4}})}^8}}}$
$\Delta E = {E_2} - {E_1}$
$
   \Rightarrow \dfrac{{0.0591}}{5}\log ({10^{ - 32}}) \\
   \Rightarrow 0.38V \\
 $

Therefore, we can say that the oxidising power of the couple decreases by $0.38V$.

Note: remember that the electrode potential is the sum of oxidising potential and reduction potential. Also remember that when two rods of the same metal are dipped separately into an electrolytic solution and they are connected by a salt bridge then the system behaves as a Galvanic cell which is also known as the concentration cell.

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