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Q.

50 mL of 0.1 M solution of a salt reacted with 25 mL of 0.1 M solution of sodium sulphite. The half reaction for the oxidation of sulphite ion is :

SO32- aq + H2Ol    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

zero

b

1

c

2

d

4

answer is C.

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Detailed Solution

No. of equivalents = mole x n-factor

SO32- + H2O    SO4-2 + 2H+  + 2e-                                                                                                                                               For Sulphite n factor is 2.       

                 (VN)salt = (VN)Sulphite                                                                                                                                                            50 × .1 × n = 25 × .1 ×2 (N=M×n-factor) n = 2.5 × 25 n = 1   Final oxidation state will be 3-1=2

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