The magnitude of the change in oxidising power of the $MnO_4^ - /M{n^{2 + }}$ couple is x $\times$ 10^$-$4 V, if the H⁺ concentration is decreased from 1M to 10^$-$4 M at 25$^\circ$C. (Assume concentration of $MnO_4^ -$ and $M{n^{2 + }}$ to be same on change in H⁺ concentration). The value of x is ___________. $\left[ {Given\,:{{2.303RT} \over F} = 0.059} \right]$

Reaction,<br/><br/>$MnO_4^ - + {H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$<br/><br/>n = 5<br/><br/>Applying Nernst equation, ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[P]} \over {[R]}}$<br/><br/>or $${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}{\left[ {{1 \over {{H^ + }}}} \right]^8}$$<br/><br/>(I) Given, [H⁺] = 1 M<br/><br/>${E_1} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}$<br/><br/>(II) Now, [H⁺] = 10^$-$4 M<br/><br/>$${E_2} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}} \times {1 \over {{{({{10}^{ - 4}})}^8}}}$$<br/><br/>$\therefore$ $\left| {{E_1} - {E_2}} \right|$<br/><br/>$$\left| {{E_1} - {E_2}} \right| = {{0.0591} \over 5} \times 32 = 0.3776\,V = 3776 \times {10^{ - 4}}$$<br/><br/>x = 3776