What will be the balanced equation in acidic medium for the given reaction ?
$C r_{2} O_{7 (a q)}^{2 -} + S O_{2} (g) \to C r_{(a q)}^{3 +} + S O_{4 (a q)}^{2 -}$

C r_{2} O_{7 (a q)}^{2 -} + 3 S O_{2} (g) + 2 H_{(a q)}^{+} \to 2 C r_{(a q)}^{3 +} + 3 S O_{4 (a q)}^{2 -} + H_{2} O (l)

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2 C r_{2} O_{7 (a q)}^{2 -} + 3 S O_{2} (g) + 4 H_{(a q)}^{+} \to 4 C r_{(a q)}^{3 +} + 3 S O_{4 (a q)}^{2 -} + 2 H_{2} O (l)

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C r_{2} O_{7 (a q)}^{2 -} + 3 S O_{2} (g) + 14 H_{(a q)}^{+} \to 2 C r_{(a q)}^{3 +} + 3 S O_{4 (a q)}^{2 -} + 7 H_{2} O (l)

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C r_{2} O_{7 (a q)}^{2 -} + 6 S O_{2} (g) + 7 H_{(a q)}^{+} \to 2 C r_{(a q)}^{3 +} + 6 S O_{4 (a q)}^{2 -} + 7 H_{2} O (l)

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Solution

The correct option is A $C r_{2} O_{7 (a q)}^{2 -} + 3 S O_{2} (g) + 2 H_{(a q)}^{+} \to 2 C r_{(a q)}^{3 +} + 3 S O_{4 (a q)}^{2 -} + H_{2} O (l)$
In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
$+ 6 C r_{2} {O_{7}}^{2 -} (a q) + + 4 S O_{2} \to + 3 C r + + 6 S O_{4}^{2 -}$
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: $+ 6 C r_{2} {O_{7}}^{2 -} (a q) \to 2 + 3 C r$
Oxidation: $+ 4 S O_{2} \to + 6 S O_{4}^{2 -}$
Step 3: oxidation-number change is:
Reduction: $+ 6 C r_{2} {O_{7}}^{2 -} \to 2 + 3 C r$ : Gain of 6 electrons total
Oxidation: $+ 4 S O_{2} \to + 6 S O_{4}^{2 -}$ Loss of total 2 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: $+ 6 C r_{2} {O_{7}}^{2 -} \to 2 + 3 C r \times 1$ : Gain of 6 electron total
Oxidation: $+ 4 S O_{2} \to + 6 S O_{4}^{2 -} \times 3$ : Loss of 6 electron
$∴$ Reduction: $+ 6 C r_{2} {O_{7}}^{2 -} \to 2 + 3 C r$
Oxidation: $3 + 4 S O_{2} \to 3 + 6 S O_{4}^{2 -}$
Step 5: Balance O atoms in reduction reaction by adding $H_{2} O$ and then balance H by $H^{+}$ as:
Reduction: $+ 6 C r_{2} {O_{7}}^{2 -} + 14 H^{+} \to 2 + 3 C r + 7 H_{2} O$
Oxidation: $3 + 4 S O_{2} + 6 H_{2} O \to 3 + 6 S O_{4}^{2 -} + 12 H^{+}$
Thus overall Balanced reaction is:
$+ 6 C r_{2} {O_{7}}^{2 -} + 14 H^{+} + 3 + 4 S O_{2} + 6 H_{2} O \to 2 + 3 C r + 7 H_{2} O + 3 + 6 S O_{4}^{2 -} + 12 H^{+}$
or $+ 6 C r_{2} {O_{7}}^{2 -} + 2 H^{+} + 3 + 4 S O_{2} \to 2 + 3 C r + H_{2} O + 3 + 6 S O_{4}^{2 -}$

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