`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?A. ZeroB. `1`C. `2`D. `4`

`50 mL` of `0.1M` solution of a salt reacted with `25 mL` of `0.1M` so

1.	`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?A. ZeroB. `1`C. `2`D. `4`
Answer» Correct Answer - C 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+)`

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