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In an experiment 50ml of 0.1(M) solution...

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?

Text Solution

Verified by Experts

The correct Answer is:
`+2`
Meq. of sodium sulphite = Meq. of salt
25 x 0.1 x 2 = 50 x 0.1 x n
`therefore n=1`
(Where n represents valence factor for metal involving no. of electrons gained)
Thus, `M^(3+) + e^(-) to M^(2+)`
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