Balance the following redox reaction in acidic medium:

$M n O_{4}^{-} + S O_{2} ⟶ S O_{4}^{2 -} + {M n}^{2 +}$

2 M n O_{4}^{-} + 5 S O_{2} + 2 H_{2} O ⟶ 5 S O_{4}^{2 -} + {2 M n}^{2 +} + 4 H^{+}

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2 M n O_{4}^{-} + 3 S O_{2} + 2 H_{2} O ⟶ 5 S O_{4}^{2 -} + {2 M n}^{2 +} + 6 H^{+}

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2 M n O_{4}^{-} + 3 S O_{2} + 2 H_{2} O ⟶ 3 S O_{4}^{2 -} + {2 M n}^{2 +} + 4 H^{+}

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None of these

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Solution

The correct option is C $2 M n O_{4}^{-} + 5 S O_{2} + 2 H_{2} O ⟶ 5 S O_{4}^{2 -} + {2 M n}^{2 +} + 4 H^{+}$
The unbalanced redox equation is as follows:
$M n O_{4}^{-} + S O_{2} ⟶ S O_{4}^{2 -} + {M n}^{2 +} (a c i d i c)$
All atoms other than H and O are balanced.
The oxidation number of Mn changes from +7 to +2. The change in the oxidation number per Mn atom is 5.
The oxidation number of S changes from +4 to +6. The change in the oxidation number is 2.
To balance the increase in the oxidation number with decrease in the oxidation number multiply $M n O_{4}^{-}$ and ${M n}^{2 +}$ with 2 and multiply $S O_{2}$ and $S O_{4}^{2 -}$ with 5.
$2 M n O_{4}^{-} + 5 S O_{2} ⟶ 5 S O_{4}^{2 -} + 2 {M n}^{2 +}$
To balance O atoms add 2 water molecules on LHS and to balance H atoms, add 4 $H^{+}$ on RHS.

$2 M n O_{4}^{-} + 5 S O_{2} + 2 H_{2} O ⟶ 5 S O_{4}^{2 -} + {2 M n}^{2 +} + 4 H^{+}$
This is the balanced chemical equation.

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Similar questions

Identify the reducing agent in the following redox reaction:

$5 S O_{3}^{2 -} + 2 M n O_{4}^{-} + 6 H^{+} \to 5 S O_{4}^{2 -} + 2 M n^{2 +} + 3 H_{2} O$

B i O_{3}^{-} + M n^{2 +} \to B i^{3 +} + M n O_{4}^{-} + H_{2} O

2 M n O_{4}^{-} + 5 H_{2} C_{2} O_{4} + 6 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

2 C u^{+}

\to

C u^{2 +}

C u^{0}

3 M n O_{4}^{2 -}

4 H^{+}

\to

2 M n O_{4}^{-}

M n O_{2}

2 H_{2} O

2 K M n O_{4}

△ - \to

K_{2} M n O_{4}

M n O_{2}

O_{2}

2 M n O_{4}^{-}

3 M n^{2 +}

2 H_{2} O

⟶

5 M n O_{2}

4 H^{+}

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

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