In an experiment, 50 mL of 0.1 M solution of a metal salt reacted with 25 mL of 0.1 M solution of sodium sulphite. The half equation for the oxidation of sulphite ion is→ S O 3 2 − ( a q ) + H 2 O

Question

In an experiment, 50 mL of 0.1 M solution of a metal salt reacted with 25 mL of 0.1 M solution of sodium sulphite. The half equation for the oxidation of sulphite ion is→

$S O_{3}^{2 -} (a q) + H_{2} O (l) \to S O_{4}^{2 -} (a q) + 2 H^{+} (a q) + 2 e^{-}$

If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal?

TestHub · Accepted Answer

M + 3 → M + n (Reduction) ​ n -factor = ( 3 − n ) e − gain ( oxidation ) n -factor = ( + 4 to + 6 ) = 2 e − loss Total loss = Total gain 0.1 M × ( 3 − n ) × 50 ml = 0.1 M × ( 2 ) × 25 ml 3 − n = 1 n = 2 ​

Chemistry - Redox Reactions Question with Solution | TestHub

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