66. In acidic medium Mno2 is an oxidant as Mno2(s) +4H(+) +2e(-) gives Mn2+ + 2H20 If pH of the solution is decreased by one unit, the electrode potential of the half cell Pt:Mno2; Mn2+ will change by

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E^{\circ}

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O i s 1.23 V F e^{3 +} + e^{-} \to F e^{2 +} i s 0.77 V

M n O_{4}^{-} / M n O_{2}

E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 v o l t, E_{M n O_{2} / M n^{2 +}}^{o} = 1.23 v o l t

△ G^{\circ}

M n O_{4}^{-} | M n O_{2}

x F

x

E_{M n O_{4}^{-} | M n^{2 +}}^{\circ} = 1.5 V; E_{M n O_{2} | M n^{2 +}}^{\circ} = 1.25 V

M n O_{4}^{⊖}

M n O_{4}^{⊖} + 8 H^{\oplus} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

H^{\oplus}

M n O_{4}^{⊖}, M n^{2 +} | P t

M n O_{4}^{-} | M n O_{2}

E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V E_{M n O_{2} / M n^{2 +}}^{0} = 1.23 V

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