Chemistry - Redox Reactions Question with Solution | TestHub
Match each solution in Column I with the solution(s) in Column II needed for complete reaction.
Column I | Column II |
|---|---|
A) 100 ml of 0.3 M - H₂C₂O₄ solution | P) 100 ml of 0.3 M - KOH solution |
B) 50 ml of 0.6 M - KHC₂O₄ solution | Q) 120 ml of 0.1 M - KMnO₄ solution in the presence of H₂SO₄ |
C) 50 ml of 0.6 M - HCl solution | R) 60 ml of 0.1 M - KMnO₄ solution in the presence of H₂SO₄ |
D) 100 ml of 0.2 M - H₃PO₄ solution | S) 100 ml of 0.6 M - KOH solution |
Options:
Answer:
Solution:
Column II equivalents are calculated as follows:
P = 0.03 mol KOH = 0.03 eq
Q = 0.012 mol KMnO₄ × 5 = 0.06 eq
R = 0.006 mol × 5 = 0.03 eq
S = 0.06 mol KOH = 0.06 eq
Let's analyze each option:
(A) For 0.03 mol H₂C₂O₄:
As an acid, it provides 0.06 eq (S).
As a reductant (C₂O₄²⁻ 2CO₂, 2 e⁻), it provides 0.06 eq (Q).
(B) For 0.03 mol KHC₂O₄:
As an acid, it provides 0.03 eq (P).
As a reductant, it provides 0.06 eq (Q).
(C) For 0.03 mol HCl:
For neutralization, it provides 0.03 eq (P).
As a reductant (2Cl⁻ Cl₂, 1 e⁻ per Cl⁻), it provides 0.03 eq (R).
(D) For 0.02 mol H₃PO₄:
Complete neutralization requires eq (S).
