Question

The equation
$\underset{\left(aq\right)}{Mn{O}_4^{-}}+\underset{\left(aq\right)}{B{r}^{-}}\to \underset{\left(s\right)}{Mn{O}_2}+\underset{\left(aq\right)}{Br{O}_3^{-}}$
when balanced in a basic medium by oxidation number method will be :

• A. $2Mn{O}_4^{-}+B{r}^{-}+H_{2}O\to 2Mn{O}_2+Br{O}_3^{-}+2O{H}^{-}$
• B. $Mn{O}_4^{-}+B{r}^{-}+H_{2}O\to 2Mn{O}_2+Br{O}_3^{-}+2O{H}^{-}$
• C. $Mn{O}_4^{-}+B{r}^{-}+H_{2}O\to Mn{O}_2+Br{O}_3^{-}+2O{H}^{-}$
• D. $2Mn{O}_4^{-}+B{r}^{-}+H_{2}O\to 2Mn{O}_2+Br{O}_3^{-}+3O{H}^{-}$

Answer 1

The correct option is A $2Mn{O}_4^{-}+B{r}^{-}+H_{2}O\to 2Mn{O}_2+Br{O}_3^{-}+2O{H}^{-}$

$Mn{O}_4^{-}+B{r}^{-}\to Mn{O}_2+Br{O}_3^{-}$






$Mn{O}_4^{-}+B{r}^{-}\to Mn{O}_2+Br{O}_3^{-}$
Oxidation state of Mn in $Mn{O}_4^{-}+7$
Oxidation state of Br in $Br{O}_3^{-}=+5$
Oxidation state of Mn in $Mn{O}_2=+4$
Oxidation state of Br in $B{r}^{-}=-1$

Cleary, $Br$ is undergoing oxidation and $Mn{O}_4^{-}$ is undergoing reduction.
using the formula of n-factor given above,
$n_f$ of $Mn{O}_4^{-}=3$
$n_f$ of $B{r}^{-}=6$
ratio of n factors =1:2
Cross multiplying these with the ratio of $n_f$'s of each other.
we get,
$2Mn{O}_4^{-}+B{r}^{-}\to Mn{O}_2+Br{O}_3^{-}$

Balancing the elements other than oxygen and hydrogen on both sides,
$2Mn{O}_4^{-}+B{r}^{-}\to 2Mn{O}_2+Br{O}_3^{-}$

Adding the $H_{2}O$ to balance the oxygen,

$2Mn{O}_4^{-}+B{r}^{-}\to 2Mn{O}_2+Br{O}_3^{-}+1H_{2}O$

Adding $H^{+}$ to balance hydrogen,

$2Mn{O}_4^{-}+B{r}^{-}+2H^{+}\to 2Mn{O}_2+Br{O}_3^{-}+1H_{2}O$

Now adding $O{H}^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2}O$,
$2Mn{O}_4^{-}+B{r}^{-}+2H^{+}+2O{H}^{-}\to 2Mn{O}_2+Br{O}_3^{-}+1H_{2}O+2O{H}^{-}$

$2Mn{O}_4^{-}+B{r}^{-}+2H_{2}O\to 2Mn{O}_2+Br{O}_3^{-}+1H_{2}O+2O{H}^{-}$
Eliminating the unrequired common $H_{2}O$ which is present on the both sides,
$2Mn{O}_4^{-}+B{r}^{-}+H_{2}O\to 2Mn{O}_2+Br{O}_3^{-}+2O{H}^{-}$
Charge balancee:
LHS : charge= -3
RHS: charge = -3
This is the final balanced equation.

Theory :
Step 1: Identifying the oxidizing/reducing agent
Step 2: find the $n_f$ by the formula $n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atom}$
Step 3: Equalising the decrease/increase in oxidation number
Cross multiply the Oxidising and the Reducing Agents by each other respectively on the reactant side.
Step 4: Balance all the atoms except O and H, without changing the stoichiometric coefficients on the reactant side
Step 5: Balancing the O atoms
Balance the O atoms by adding $H_{2}O$
Step 6:
As soon as we add $H_{2}O$, we add twice the $H^{+}$ ions on the opposite side.
$x{H}_{2}O⟶2x{H}^{+}$

For Basic medium :
1. Add to both sides the same number of $O{H}^{-}$ and $H^{+}$
2. Combine $H^{+}$ and $O{H}^{-}$ to form $H_{2}O$
3. Cancel the equal number of $H_{2}O$ molecules on both sides whenever possible.