Match each process in Column I with the equivalent weight of Cl₂ (molar mass 71) in Column II.
Column I
Column II
A) Cl₂ $\to$ Cl⁻
P) 71
B) Cl₂ $\to$ ClO₃⁻
Q) 35.5
C) Cl₂ $\to$ Cl⁻ + ClO₃⁻
R) 42.6
D) Cl₂ $\to$ Cl⁻ + ClO⁻
S) 7.1

Question

Match each process in Column I with the equivalent weight of Cl₂ (molar mass 71) in Column II.

Column I	Column II
A) Cl₂ $\to$ Cl⁻	P) 71
B) Cl₂ $\to$ ClO₃⁻	Q) 35.5
C) Cl₂ $\to$ Cl⁻ + ClO₃⁻	R) 42.6
D) Cl₂ $\to$ Cl⁻ + ClO⁻	S) 7.1

TestHub · Accepted Answer

Equivalent weight = 71 / (electrons transferred per Cl₂). (A) Cl(0) → -1: 2 e⁻ per Cl₂, E = 71/2 = 35.5 (Q). (B) Cl(0) → +5: 10 e⁻ per Cl₂, E = 71/10 = 7.1 (S). (C) Disproportionation to Cl⁻ and ClO₃⁻: for a disproportionating species the overall equivalent weight is the sum of the two half-reaction values, 35.5 + 7.1 = 42.6 (R). (D) Cl₂ → ClO⁻ is Cl(0) → +1, 2 e⁻, E = 35.5; disproportionation sum = 35.5 + 35.5 = 71 (P).

Chemistry - Redox Reactions Question with Solution | TestHub

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