Redox Reactions — Chemistry STD 11 Science — Question

$\text{MnO}_4^-\text{(aq)}+\text{Br}\text{(aq)}\xrightarrow{ \ \ \ \ \ }\text{MnO}_2\text{(aq)}+\text{BrO}_3^-\text{(aq)}\\ \ +7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +5$

Step 2: $\text{MnO}_4^-$ is oxidant because its oxidation state is decreasing. Br^- is reductant because its oxidation state is increasing.

Step 3: Oxidation state of Mn is decreasing by 3. Oxidation state of Br is increasing by 6. To equalize increase and decrease, multiply $\text{MnO}_4^-$ by 2 and Br^- by 1 we get.

$2\text{MnO}_4^-\text{(aq)}+\text{Br}^-\text{(aq)}\xrightarrow{ \ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{Br}_2^-\text{(aq)}$

Step 4: Now for balncaing oxygen, we add 1 molecule of H₂O on RHS.

$2\text{MnO}_4^-\text{(aq)}+\text{Br}^-\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{BrO}_3^-\text{(aq)}+\text{H}_2\text{O}\text{(l)}$

Step 5: As the reaction is taking place in basic medium to balance hydrogen, add 2H₂O molecules on LHS and 2OH^- on RHS.

$2\text{MnO}_4^-\text{(aq)}+\text{Br}^-\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{BrO}_3^-\text{(aq)}+\text{H}_2\text{O}\text{(l)}+2\text{OH}^-\text{(aq)}$

It can be seen 1 molecule of H₂O gets cancelled on both sides, we get.

$2\text{MnO}_4^-\text{(aq)}+\text{Br}^-\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{BrO}_3^-\text{(aq)}+\text{H}_2\text{O}\text{(l)}+2\text{OH}^-\text{(aq)}$

is a balanced equation.