The following is a partial redox equation:
$M n O_{4}^{-} + 8 H^{+} + x e^{-} ⟶ M n^{2 +} + 4 H_{2} O$

The value of the coefficient 'x' is:

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Solution

The correct option is A
5

When you account for the change in the oxidation number of Manganese on either side:
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O$
x = 5.

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Similar questions

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

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

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

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^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{o} = X_{1} V

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{o} = X_{2} V

E^{o}

M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O

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