The magnitude of the change in oxidising power of the MnO _ 4 ^ - /...

Eqn is-

${MnO}_{4}^{-} + {H}^{⊕} + 5{e}^{-} \to {Mn}^{+ 2} + 4{H}_{2}O$

Nernst equation:

${E}_{cell} = {E}_{Cell}^{0} - \dfrac{0 . 059}{5}log\dfrac{\left[{Mn}^{+ 2}\right]}{\left[{MnO}_{4}^{-}\right]}{\left[\dfrac{1}{{H}^{+}}\right]}^{8}$

(I) Given \left[{H}^{⊕}\right] = 1M$</p><p>${E}_{1} = {E}^{0} - \dfrac{0 . 059}{5}log\dfrac{\left[{Mn}^{+ 2}\right]}{\left[{MnO}_{4}^{-}\right]}$</p><p>(II) Now: \left[{H}^{⊕}\right] = {10}^{- 4}M$

${E}_{2} = {E}^{0} - \dfrac{0 . 059}{5}log\dfrac{\left[{Mn}^{+ 2}\right]}{\left[{MnO}_{4}^{-}\right]} \times \dfrac{1}{{\left({10}^{- 4}\right)}^{8}}$

$= {E}^{0} - \dfrac{0 . 059}{5}log\dfrac{{Mn}^{+ 2}}{\left[{MnO}_{4}^{-}\right]} + \dfrac{0 . 059}{5}log{10}^{- 32}$

therefore: \left|{E}_{1} - {E}_{2}\right| = \dfrac{0 . 059}{5} \times 32$</p><p>$= 0 . 3776V = 3776 \times {10}^{- 4}$</p><p>$x = 3776