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 sulfite. The half-equation for the oxidation of sulfite 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 sulfite. The half-equation for the oxidation of sulfite 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 the metal in the salt was 3, what would be the new oxidation number of the metal?

TestHub · Accepted Answer

The reaction is M + 3 → M + n (Reduction). n-factor for reduction is ( 3 − n ) e − gain. For oxidation, the n-factor is ( + 4 t o + 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

Doubts & Discussion

JEE Main + Advanced (1 year 12th class)