Match each redox reaction in Column I with the mole ratio of oxidizing agent to reducing agent in Column II.
Column I
Column II
A) MnO₄⁻ + C₂O₄²⁻ $\to$ MnO₂ + CO₂
P) 2:1
B) ClO⁻ + Fe(OH)₃ $\to$ Cl⁻ + FeO₄²⁻
Q) 3:1
C) HO₂⁻ + Cr(OH)₃⁻ $\to$ CrO₄²⁻ + HO⁻
R) 2:3
D) N₂H₄ + Cu(OH)₂ $\to$ N₂O + Cu
S) 3:2

Question

Match each redox reaction in Column I with the mole ratio of oxidizing agent to reducing agent in Column II.

Column I	Column II
A) MnO₄⁻ + C₂O₄²⁻ $\to$ MnO₂ + CO₂	P) 2:1
B) ClO⁻ + Fe(OH)₃ $\to$ Cl⁻ + FeO₄²⁻	Q) 3:1
C) HO₂⁻ + Cr(OH)₃⁻ $\to$ CrO₄²⁻ + HO⁻	R) 2:3
D) N₂H₄ + Cu(OH)₂ $\to$ N₂O + Cu	S) 3:2

TestHub · Accepted Answer

(A) Mn(+7) → MnO₂ (+4) gains 3 e⁻; C₂O₄²⁻ loses 2 e⁻. LCM 6: 2 MnO₄⁻ : 3 C₂O₄²⁻, ratio 2:3 (R). (B) Cl(+1) → Cl⁻ gains 2 e⁻; Fe(+3) → +6 loses 3 e⁻. LCM 6: 3 ClO⁻ : 2 Fe(OH)₃, ratio 3:2 (S). (C) HO₂⁻ (O at -1) → OH⁻ (O at -2) gains 2 e⁻ per HO₂⁻; Cr(+3) → +6 loses 3 e⁻. LCM 6: 3 HO₂⁻ : 2 Cr, ratio 3:2 (S). (D) In N₂H₄ both N (-2) go to N₂O (avg +1), losing 6 e⁻ per N₂H₄; Cu(+2) → Cu(0) gains 2 e⁻. LCM 6: 3 Cu(OH)₂ : 1 N₂H₄, ratio 3:1 (Q).

Chemistry - Redox Reactions Question with Solution | TestHub

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