Question

Both oxidation and reduction takes place in :

• A. $NaBr+HCl⟶NaCl+HBr$
• B. $HBr+AgN{O}_3⟶AgBr+HN{O}_3$
• C. $H_2+B{r}_2⟶2HBr$
• D. $CaO+H_{2}S{O}_4\to CaS{O}_4+H_{2}O$

Answer 1

Answer: (C)

In the reaction :
$H_2+B{r}_2\to 2HBr$
Oxidation number of $H$ in $H_2$ is zero whereas its value is $+1$ in $HBr$ ; similarly oxidation number of $Br$ in $B{r}_2$ is zero where as its value is $-1$ in $HBr$. So, here $H$ is oxidised and $Br$ reduced. In all other reactions, there in no change in the oxidation number of any element.