In an experiment `50ml` of `0.1(M)` solution of a salt is reacted with `25ml` of `0.1(M)` solution of sodium sulphite. The half equation for the oxidation of sulphite ion is `SO_(3)^(2-)(aq)+H_(2)O rarr SO_(4)^(2-)(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

Let now `ON` of metal `=x`
`SO_(3)^(2-) rarr sO_(4)^(2-) x-factor=2`
milliequiv of salt=milliequiv of sodium sulphite
`50xx0.1(3-x)=25xx0.1xx2`
or `3-x=1`
`x=3-1=2`