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Question

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:
SO23(aq)+H2OSO24(aq)+2H+(aq)+2e
If the oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal?

A
0.0
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B
1
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C
2
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D
4
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Solution

The correct option is C 2
SO23 get oxidised to SO24 and release 2 electrons so its 'n' factor is 2.
Thus, the metal must have been reduced.
No. of equivalents = No. of moles × n-factor
Applying the law of equivalence,
50×0.1×n=25×0.1×2
n=1
As the metal is reduced, new oxidation number = 3n = 31 = 2.

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