`50 mL` of `0.1 M` solution of a salt reacted with `25 mL` of `0.1 M` solution of sodium sulphite. The half reaction for the oxidation of sulphite ion is:
`SO_(3)^(2-)(aq)+H_(2)O(l) rarr (aq)+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. 0
B. 1
C. 2
D. 4

Answer

C

Explanation

`SO_(3)^(2-)+H_(2)OtoSO_(4)^(2-)+2H^(o+)+2e^(-)` ltBrgt `m" Eq of "salt =m" Eq of "SO_(3)^(2-)`
`implies(0.1xxn)xx50-=(0.1xx2)xx25`
`impliesn=1`
`M^(3+)+e^(-)toM^(2+)`