Question

$50$ mL of $0.1M$ solution of a salt reacted with $25$ mL of $0.1M$ solution of sodium sulphite. The half reaction for the oxidation of sulphite ion is,

$S{O}_3^{2-}\left(aq\right)+H_{2}O\left(\ell \right)⟶S{O}_4^{2-}\left(aq\right)+2H^{+}\left(aq\right)+2e^{-}$.

If the oxidation number of metal in the salt was $3$, what would be the new oxidation number of metal?

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

$2$

Number of Equivalents of salt $=$ Number of Equivalents of sodium sulphite.

From the oxidation half reaction, $S{O}_3^{2-}\left(aq\right)+H_{2}O\left(\ell \right)⟶S{O}_4^{2-}\left(aq\right)+2H^{+}\left(aq\right)+2e^{-}$

$n$-factor of sodium sulphite $=2$.

Let, the $n$-factor of salt be $n$.

No. of equivalents $=$ No. of moles $\times$ n-factor

$50\times 0.1\times n=25\times 0.1\times 2$

$n=1$

As the metal is reduced, new oxidation number $=3-n$ $=2$.

Hence, the correct option is $C$