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Equivalent weight of ${SO}_{2}$ as reducing agent is

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answer is D.

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Detailed Solution

A reducing agent undergoes oxidation

Ox.state of sulphur in ${SO}_{2}$ = +4

Due to oxidation its oxidation number increases to '+6'

\mathop S\limits^{ + 4} {O_2} \to \mathop S\limits^{ + 6} O_4^{ - 2}

EW of\; SO_2 (as\; reductant) = \frac{64}{2}=32

Ex:-

[{Fe}^{+ 3} + e^{-} \to {Fe}^{+ 2}] \times 2

SO_2+2H_2O\rightarrow SO_4^{-2}+4H^++2e^-

2F{e^{ + 3}} + S{O_2} + 2{H_2}O \to 2F{e^{ + 2}} + SO_4^{ - 2} + 4{H^ + }

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