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Q.

In an experiment 50 ml of 0.1 (M) solution of a salt is reacted with 25 ml of 0.1 (M) solution of sodium sulphite. The half equation for the oxidation of sulphite ion isSO32-aq + H2Ol →  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 ON of metal = xSO32- →  SO42-  x-factor = 2milliequiv of salt = milliequiv of sodium sulphite50 × 0.1 3-x = 25 × 0.1 ×2 or  3 - x = 1 x = 3 -1 = 2
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