In the basic disproportionation reaction , the equivalent weight of Bromine () is mathematically exp — Redox Reactions and Volumetric Analysis Chemistry Question
Question
In the basic disproportionation reaction $Br_2 + KOH \rightarrow KBr + KBrO_3 + H_2O$, the equivalent weight of Bromine ($Br_2$) is mathematically expressed as $\frac{A \cdot M}{B}$ (where M is its molar mass, and A/B is the simplest irreducible fractional ratio). Find the value of $(A + B)$.
Answer: 8
💡 Solution & Explanation
$Br_2 (0) \rightarrow 2Br^- (-1)$. Gain of 2e- (Reduction n-factor, $n_1 = 2$). $Br_2 (0) \rightarrow 2BrO_3^- (+5)$. Loss of 10e- (Oxidation n-factor, $n_2 = 10$). For disproportionation, $E = M/n_1 + M/n_2 = M/2 + M/10 = 6M/10 = 3M/5$. Therefore, $A = 3$ and $B = 5$. The sum $(A + B) = 3 + 5 = 8$.
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