The value of x in the following reaction- MnO-4-8H-xe- Mn2-4H2O is-

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Ion Electron Method

The value of ...

Question

The value of x in the following reaction, $M n O_{4}^{-} + 8 H^{+} + x e \to M n^{2 +} + 4 H_{2} O$ is?

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M n O_{4}^{-} + 8 H^{+} + x e = M n^{2 +} + 4 H_{2} O

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{O} = 1.51 V; Δ G_{1}^{0} = - 5 \times 1.51 \times F

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{O} = 1.23 V; Δ G_{2}^{0} = - 5 \times 1.23 \times F

E_{M n O_{4}^{-} | M n O_{2}}^{\circ}

5 {F e}^{2 +} + M n O_{4}^{-} + 8 H^{+} ⇌ {M n}^{2 +} + 5 {F e}^{3 +} + 4 H_{2} O

{F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}, {E_{1}}^{o}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ {M n}^{2 +} + 4 H_{2} O, {E_{2}}^{o}

M n O_{4}^{-} + 8 H^{+} + x e^{-} ⟶ M n^{2 +} + 4 H_{2} O

n

M n O_{4}^{-} + 8 H^{+} + n e^{-} ⟶ M n^{2 +} + 4 H_{2} O

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