4) In an experiment, 50 mL of 0 1 M solution of a salt reacted with 25 mL of - Chemistry - General Principles and Processes of Isolation of Elements

Question

4) 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 is S O   ( a q )   + H   O
  → S O   ( a q )   + 2 H   + 2 e

Geetha · Accepted Answer

The oxidation state of the sulphur inSO32-x + 3(-2) = -2x-6 = -2x= +4SO42-x + 4(-2) = -2x-8 = -2x= +6Thus the n factor is 2
As the oxiation number is increased S undergoes oxidation
 So the metal is reduced
The milli mole equivalent of the salt isBefore oxidation = Volume × Molarity =50 × 0
1 = 5 milli moleAfter oxidation =  Volume × Molarity =25 × 0
1 = 2
5 milli moleApplying the law of equivalence50 × 0
1 × (3-n) = 25 × 0
1 × 215-5n=5-5n = -105n= 10n= 2 Thus the oxidation state of the metal is 2

4) 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 is S O ( a q ) + H O → S O ( a q ) + 2 H + 2 e

4) 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 is $S O (a q) + H O \to S O (a q) + 2 H + 2 e$