`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 sulphate ion is:
`SO_(3(aq.))^(2-)+H_(2)O_((l))rarrSO_(4(aq.))^(2-)+2H_(+(aq.))+2e`
If the oxidation number of metal in the salt was `3`, what would be the new oxidation number of metal?

Options

A. Zero
B. `1`
C. `2`
D. `4`

Answer

C

Explanation

Meq.of sodium sulphate`=` Meq.of salt
`25xx0.1xx2=50xx0.1xxn :. n=1` (where `n` represents valence factor for metal involving no of electrons gained)
Thus `M^(3+)+erarrM^(2+)`