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In an experiment, 50 ml. of 0. 1 (M) solution of a salt reacted with 25 ml. of 0.1 (M) solution of sodium sulphite. The half equation for the oxidation of sulphite ion is -SO32−(aq)+H2O(λ)⟶SO42−(aq)+2H+(aq)+2e−If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal?

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a

0

b

1

c

2

d

4

answer is C.

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Detailed Solution

Let new O.N of metal be xSO32−→SO42−x -factor =250×0.1(3−x)=25×0.1×2 or 3−x=1⇒x=3−1=2

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In an experiment, 50 ml. of 0. 1 (M) solution of a salt reacted with 25 ml. of 0.1 (M) solution of sodium sulphite. The half equation for the oxidation of sulphite ion is -SO32−(aq)+H2O(λ)⟶SO42−(aq)+2H+(aq)+2e−If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal?