Question

How to balance the following equation by oxidation number method.
$B{r}^{-}+Mn{O}_4^{-}\to Br{O}_3^{-}+Mn{O}_2$

Answer 1

Step 1: The skeletal ionic equation is
$Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)\to Mn{O}_2\left(s\right)+Br{O}_3^{-}\left(aq\right)$
Step 2: Assign oxidation numbers for Mn and Br
$\begin{array}{cccc}+7& -1& +4& +5\\ Mn{O}_4^{-}\left(aq\right)& B{r}^{-}\left(aq\right)& Mn{O}_2\left(s\right)& Br{O}_3^{-}\left(aq\right)\end{array}$
Step 3: Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease.
$2Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)\to 2Mn{O}_2\left(s\right)+Br{O}_3^{-}\left(aq\right)$
Step 4: As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add $2O{H}^{-}$ ions on the right to make ionic charges equal.
$2Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)\to 2Mn{O}_2\left(s\right)+Br{O}_3^{-}\left(aq\right)+2O{H}^{-}$
Step 5: Finally, count the hydrogen atoms and add appropriate number of water molecules (i.e., one $H_{2}O$ molecules) on the left side to achieve balanced redox change.
$2Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)+H_{2}O\to 2Mn{O}_2\left(s\right)+Br{O}_3^{-}\left(aq\right)+2O{H}^{-}$