Balance the following redox reactions:

${S O}_{2} (g) + H {N O}_{2} (a q) ⟶ H_{2} {S O}_{4} (a q) + N O (g)$

{S O}_{2} (g) + 2 H {N O}_{2} (a q) ⟶ H_{2} {S O}_{4} (a q) + 2 N O (g)

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{S O}_{2} (g) + 3 H {N O}_{2} (a q) ⟶ H_{2} {S O}_{4} (a q) + 2 N O (g)

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{S O}_{2} (g) + 2 H {N O}_{2} (a q) ⟶ H_{2} {S O}_{4} (a q) + 4 N O (g)

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3 {S O}_{2} (g) + 2 H {N O}_{2} (a q) ⟶ H_{2} {S O}_{4} (a q) + 2 N O (g)

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Solution

The correct option is A ${S O}_{2} (g) + 2 H {N O}_{2} (a q) ⟶ H_{2} {S O}_{4} (a q) + 2 N O (g)$
The balanced reaction is given below:
${S O}_{2} (g) + 2 H {N O}_{2} (a q) ⟶ H_{2} {S O}_{4} (a q) + 2 N O (g)$

S is getting oxidized from $+ 4$ to $+ 6$ .
Hence, there is a transfer of $2$ electrons.
The factor of $2$ will precede $H N O_{2}$ and $N O$ .

N is getting reduced from $+ 3$ to $+ 2$ .
Hence, there is a transfer of $1$ electron.
The factor of $1$ will precede $S O_{2}$ and $H_{2} S O_{4}$ .

Hence, option A is correct.

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