What is the standard electrode potential for the electrode MnO4 - - MnO2 in solution - EoMnO4- - Mn2- - 1-51 volt- EoMnO2-Mn2- - 1-23 volt

The correct option is C 1.70 V MnO_4^ + 8H^+ + 5e^ #160;→ #160; #160;Mn^2+ + 4 H_2O ; E^o = 1.51 #160;volt MnO_2 + 4H^+ + 2e^ → Mn^2+ ...

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What is the s...

Question

What is the standard electrode potential for the electrode $M n O_{4}^{-} / M n O_{2}$ in solution : $E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 v o l t, E_{M n O_{2} / M n^{2 +}}^{o} = 1.23 v o l t$

0.7 V

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3.33 V

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1.70 V

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2.5 V

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Similar questions

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

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

M n O_{2} / M n^{2 +} = 1.23 V

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

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

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

E_{M n O_{2} / M n^{2 +}}^{o} = 1.23 V

M n O_{4}^{-}

0.8 M H^{+}

F e^{2 +}

90 %

M n O_{4}^{-}

M n^{2 +}

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

V

\frac{M n O_{4}^{-}}{M n O_{2}}

E^{\circ} M n_{4}^{O -}, M n^{2 +} = 1.51 V E^{\circ} M n O_{2}^{-}, M n^{2 +} = 1.23 V

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