Question

When the following equation $B{r}_2+S{O}_2+H_{2}O\to H_{2}S{O}_4+HBr$ is balanced, the coefficient, in the lowest whole number of $HBr$ is :

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Answer 1

The correct option is B 2

$\begin{array}{c}\text{Given is a redox reaction, in which,}\\ B{r}^0\text{goes to}B{r}^{-}\text{and}{s}^{+4}\text{goes to}{s}^{+6}\text{.}\\ \text{thence,}B{r}_2\text{is acting as oxidising agent while}\\ S{O}_2\text{is acting as reducing agent.}\\ \text{Note that,}{\left(n_f\right)}_{OA}=2\text{and}{\left(n_f\right)}_{RA}=2\text{.}\end{array}$

$\text{Thus,}2{Br}_2+2{SO}_2+4H_{2}O⟶2H_2{SO}_4+4HBr$

$\text{Simplified}:B{r}_2+S{O}_2+2H_{2}O⟶H_{2}S{O}_4+2HBr$

Option B is correct answer