Question

In an experiment, 50 mL of 0.1 M solution of a salt reacted with 25 mL of 0.1 M solution of sodium sulphite. The half equation for the oxidation of sulphite ion is:
$S{O}_{3\left(aq\right)}^{2-}+H_{2}O\to S{O}_{4\left(aq\right)}^{2-}+2H_{\left(aq\right)}^{+}+2e^{-}$
If the oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal?

• A. \N
• B. 1
• C. 2
• D. 4

Answer 1

The correct option is C 2

$S{O}_3^{2-}$ get oxidised to $S{O}_4^{2-}$ and release 2 electrons so its 'n' factor is 2.
Thus, the metal must have been reduced.
No. of equivalents = No. of moles × n-factor
Applying the law of equivalence,
$50\times 0.1\times n=25\times 0.1\times 2$
$n=1$
As the metal is reduced, new oxidation number = $3-n$ = $3-1$ = 2.