$F e^{2 +} \to F e^{3 +} + e^{-}; M n O_{4}^{-} + 5 e^{-} \to M n^{2 +}$ , the ratio of stoichiometric coefficient of $F e^{2 +}$ and $M n O_{4}^{-}$ in the balanced redox reaction is:

1 : 5

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5 : 1

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2 : 3

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6 : 1

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Solution

The correct option is C 5 : 1
$F e^{2 +} \to F e^{3 +} + e^{-}$ ------ 1 electron is lost in the reaction
$M n O_{4}^{-} + 5 e^{-} \to M n^{2 +}$ ----- 5 electrons are gained in this reaction
So the ratio of the stoichiometric coefficient of $F e^{2 +}$ and $M n O_{4}^{-}$ is $5 : 1$ when the redox reaction occurs between them.
$5 F e^{2 +} + M n O_{4}^{-} + 8 H^{+} \to 5 F e^{3 +} + M n^{2 +} + 4 H_{2} O$

Hence option B is correct.

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Similar questions

F e S + M n O_{4}^{-}^{-} F e^{3 +} + S O_{4}^{2 -} + M n^{2 +}

M n O_{4}^{-}

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

M n O_{4}^{-} + a F e^{2 +} + b H^{+} \to M n^{2 +} + c F e^{3 +} + d H_{2} O

M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} ⟶ M n^{2 +} + C O_{2} + H_{2} O

M n O_{4}^{-}, C_{2} O_{4}^{2 -} a n d H^{+}

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

M n O_{4}^{-}, C_{2} O_{4}^{3 +}

H^{+}

M n O_{4}^{-} + C_{2} O_{4}^{2 -} \to M n^{2 +} + C O_{2}

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