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Valency and Oxidation State

In an experim...

Question

In an experiment, $50$ mL of $0.1$ M solution of a metallic salt reacted exactly with $25$ mL of $0.1$ M solution of sodium sulphite. In the reaction, ${S O_{3}}^{-}$ is oxidised 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?

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2.5

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Solution

The correct option is C 2.5
Total number of millimoles of metallic salt present $= 50 m L \times 0.1 M = 5$ millimoles.
Total number of millimoles of sodium sulphite $= 25 m L \times 0.1 M = 2.5$ millimoles.
The oxidation number of S changes from +5 to +6.
Increase in the oxidation number of 1 S atom = 1.
Total number of millimoles electrons lost by sodium sulphite = 2.5
5 millimoles of metallic salt gains 2.5 millimoles of electrons.
1 molecule of metallic salt will gain 0.5 electron.
Hence, the oxidation number of metal will decrease from 3 to 2.5.

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