Question

In an experiment, $50mL$ of $0.1M$ solution of a metallic salt reacted exactly with $25mL$ of $0.1M$ solution of sodium sulphite. In the reaction $S{O}_3^{2-}$ is oxidized to $S{O}_4^{2-}$. If the original oxidation number of the metal in the salt was 3 , what would be the new oxidation number of the metal?

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (C)

$S{O}_3^{2-}$ is oxidized to $S{O}_4^{2-}($ change in $O.N.=2)$
$25mL$ of $0.1MS{O}_3^{2-}=2.5$ millimol
$=5.0$ milliequivalents of $S{O}_3^{2-}$
$=5.0$ milliequivalents of $M^{3+}$
$50mL$ of $0.1M{M}^{3+}=5$ millimol (given)
Thus, decrease in O.N. of $M^{3+}$ should be $1$
So that $5$ millimol $=5$ milliequivalents
Thus, new O.N. of metal $=2$