Match each process in Column I with the equivalent weight of $B r_{2}$ (molar mass 160) in Column II.
Column I
Column II
A) $B r_{2}$ $\to$ $B r^{-}$
P) 160
B) $B r_{2}$ $\to$ $Br O_{3}^{-}$
Q) 80
C) $B r_{2}$ $\to$ $B r^{-}$ + $Br O_{3}^{-}$
R) 96
D) $B r_{2}$ $\to$ $B r^{-}$ + $Br O^{-}$
S) 16

Question

Match each process in Column I with the equivalent weight of $B r_{2}$ (molar mass 160) in Column II.

Column I	Column II
A) $B r_{2}$ $\to$ $B r^{-}$	P) 160
B) $B r_{2}$ $\to$ $Br O_{3}^{-}$	Q) 80
C) $B r_{2}$ $\to$ $B r^{-}$ + $Br O_{3}^{-}$	R) 96
D) $B r_{2}$ $\to$ $B r^{-}$ + $Br O^{-}$	S) 16

TestHub · Accepted Answer

Equivalent weight = 160 / (electrons transferred per B r 2 ​ ). (A) Br(0) → -1: 2 e − per B r 2 ​ , E = 160/2 = 80 (Q). (B) Br(0) → +5: 10 e − per B r 2 ​ , E = 160/10 = 16 (S). (C) Disproportionation to B r − and Br O 3 − ​ : sum of the two half-reaction values, 80 + 16 = 96 (R). (D) B r 2 ​ → Br O − is Br(0) → +1, 2 e − , E = 80; disproportionation sum = 80 + 80 = 160 (P).

Chemistry - Redox Reactions Question with Solution | TestHub

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