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Balancing Redox Equations

For the redox...

Question

For the redox reaction,

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

The correct coefficients of the reactants for the balanced equation are:

M n O_{4}^{-}

- 2,

C_{2} O_{4}^{2 -}

- 16,

H^{+}

- 5

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M n O_{4}^{-}

- 2,

C_{2} O_{4}^{2 -}

- 5,

H^{+}

- 16

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M n O_{4}^{-}

- 16,

C_{2} O_{4}^{2 -}

- 5,

H^{+}

- 2

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M n O_{4}^{-}

- 5,

C_{2} O_{4}^{2 -}

- 16,

H^{+}

- 2

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Solution

The correct option is B $M n O_{4}^{-}$ - 2, $C_{2} O_{4}^{2 -}$ - 5, $H^{+}$ - 16

n - factor of $M n O_{4}^{-} \Rightarrow 5$
n - factor of $C_{2} O_{4}^{2 -} \Rightarrow 2$
Ration of n - factors of $M n O_{4}^{-}$ and $C_{2} O_{4}^{2 -}$ is 5 : 2
So, molar ratio in balanced reaction is 2 : 5
$∴$ The balanced equation is
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$

Suggest Corrections

Similar questions

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

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

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

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

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