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Question: If the \({H^ + }\) concentration decreased from \(1M\) to \({10^{ - 4}}M\) at \({25^\circ }C\) for t...

If the H+{H^ + } concentration decreased from 1M1M to 10−4M{10^{ - 4}}M at 25∘C{25^\circ }C for the couple MnO−4Mn{O^{ - 4}}/Mn2+M{n^{2 + }}, then the oxidising power of the MnO−4Mn{O^{ - 4}}/Mn2+M{n^{2 + }} couple decreases by how much?

Explanation

Solution

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:
MnO4−+8H++5e−→Mn2++4H2OMnO_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:
Ecell=E∘−0.0591nlog⁡Kc{E_{cell}} = {E^\circ } - \dfrac{{0.0591}}{n}\log {K_c}, where nn is the number of electrons.
We are given the concentration of MnO−4Mn{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 MnO−4Mn{O^{ - 4}} when the concentration is 1M1M, we will get:
E1=E∘−0.05915log⁡[Mn2+](1)8{E_1} = {E^\circ } - \dfrac{{0.0591}}{5}\log \dfrac{{[M{n^{2 + }}]}}{{{{(1)}^8}}}
Similarly, the final electrode potential will be given as:
E2=E∘−0.05915log⁡[Mn2+](10−4)8{E_2} = {E^\circ } - \dfrac{{0.0591}}{5}\log \dfrac{{[M{n^{2 + }}]}}{{{{({{10}^{ - 4}})}^8}}}
ΔE=E2−E1\Delta E = {E_2} - {E_1}
⇒0.05915log⁡(10−32) ⇒0.38V  \Rightarrow \dfrac{{0.0591}}{5}\log ({10^{ - 32}}) \\\ \Rightarrow 0.38V \\\

Therefore, we can say that the oxidising power of the couple decreases by 0.38V0.38V.

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.