Question

Consider the following redox reaction:
$Mn\left(OH{)}_2\right(s)+Mn{O}_4^{-}(aq)\to Mn{O}_2(s)$
If $x,y,z$ are the stoichiometric coefficients of $Mn(OH{)}_2,Mn{O}_4^{-}$ and $Mn{O}_2$ respectively when the above reaction is balanced in basic medium.
What will be the value of $x+y+z$ ?

• A. 5
• B. 8
• C. 10
• D. 12

Answer 1

The correct option is C 10

$Mn\left(OH{)}_2\right(s)+Mn{O}_4^{-}(aq)\to Mn{O}_2(s)$

Mn oxidation state in $Mn(OH{)}_2=+2$
Mn oxidation state in $Mn{O}_4^{-}=+7$
Mn oxidation state in $Mn{O}_2=+4$

Cleary, $Mn(OH{)}_2$ is undergoing oxidation and $Mn{O}_4^{-}$ is undergoing reduction.
Formula used for the n-factor calculation,
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atoms}$
using the formula of n-factor given above,
$n_f$ of $Mn(OH{)}_2=2$
$n_f$ of $Mn{O}_4^{-}=3$
Cross multiplying these with $n_f$ of each other.
we get,
$3Mn\left(OH{)}_2\right(s)+2Mn{O}_4^{-}(aq)\to Mn{O}_2(s)$

Balancing the Mn element on both sides,
$3Mn\left(OH{)}_2\right(s)+2Mn{O}_4^{-}(aq)\to 5Mn{O}_2(s)$

Adding the $H_{2}O$ to balance the oxygen,
The LHS of the equation contains 14 oxygens while RHS contains only 10 oxygens, so adding 4 $H_{2}O$ to RHS.
$3Mn\left(OH{)}_2\right(s)+2Mn{O}_4^{-}(aq)\to 5Mn{O}_2(s)+4H_{2}O$


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

As LHS of the equation contains 6 hydrogens while RHS contains 8 hydrogens, so adding 2$H^{+}$ to LHS.
$3Mn\left(OH{)}_2\right(s)+2Mn{O}_4^{-}(aq)+2H^{+}\to 5Mn{O}_2(s)+4H_{2}O$

Now adding $O{H}^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2}O$,
$3Mn\left(OH{)}_2\right(s)+2Mn{O}_4^{-}(aq)+2H^{+}+2O{H}^{-}\to 5Mn{O}_2(s)+4H_{2}O+2O{H}^{-}$
$3Mn\left(OH{)}_2\right(s)+2Mn{O}_4^{-}(aq)+2H_{2}O\to 5Mn{O}_2(s)+4H_{2}O+2O{H}^{-}$

Eliminating the unrequired common $H_{2}O$ which is present on the both sides,

$3Mn\left(OH{)}_2\right(s)+2Mn{O}_4^{-}(aq)\to 5Mn{O}_2(s)+2H_{2}O+2O{H}^{-}$
Charge is also equal on bth sides and is $-2$
So, this is the final balanced equation.

$x=3y=2z=5x+y+z=10$