The value of $n$ in $M n O_{4}^{-} + 8 H^{+} + n e^{-} ⟶ M n^{2 +} + 4 H_{2} O$ is:

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Solution

The correct option is A 5
${M n_{4}}^{-} + 8 H^{+} + n e^{-} \to M n^{2 +} + 4 H_{2} O$

$\Rightarrow - 1 + 8 - n = + 2$

$\Rightarrow n = + 5$

Hence, option A is correct.

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Similar questions

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}

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

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

E^{\circ} = 1.51 V

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

E^{\circ} = 1.23 V

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}

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

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