Indicate reducing agent in following reactions. $5 H_{2} O_{2} + B r_{2} ⟶ 2 H B r O_{3} + 4 H_{2} O$

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Solution

${5 H}_{2} {- 1 O}_{2} + {0 B r}_{2} ⟶ 2 H + 7 B r O_{3} + 4 H_{2} - 2 O$
Here, $H_{2} {- 1 O}_{2}$ is reduced to $H_{2} - 2 O$
& ${0 B r}_{2}$ is oxidised to $H + 7 B r O_{3}$
Therefore, ${0 B r}_{2}$ acts as a reducing agent to reduce $H_{2} - 1 O$ to $H_{2} - 2 O$ by itself getting oxidized to $H + 7 B r O_{3}$ .

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