Using Oxidation-number method, find the net ionic equation for the following reaction in basic medium:
$M n (O H)_{2} (s) + M n O_{4}^{-} (a q) \to M n O_{2} (s)$

6 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + O H^{-}

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4 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 4 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}

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2 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}

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3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}

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Solution

The correct option is D $3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}$
$M n (O H)_{2} (s) + M n O_{4}^{-} (a q) \to M n O_{2} (s)$

Mn oxidation state in $M n (O H)_{2} = + 2$
Mn oxidation state in $M n O_{4}^{-} = + 7$
Mn oxidation state in $M n O_{2} = + 4$

Cleary, $M n (O H)_{2}$ is undergoing oxidation and $M n O_{4}^{-}$ is undergoing reduction.
Formula used for the n-factor calculation,
$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atoms$
using the formula of n-factor given above,
$n_{f}$ of $M n (O H)_{2} = 2$
$n_{f}$ of $M n O_{4}^{-} = 3$
Cross multiplying these with $n_{f}$ of each other.
we get,
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to M n O_{2} (s)$

Balancing the Mn element on both sides,
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s)$

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.
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 4 H_{2} O$

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

As LHS of the equation contains 6 hydrogens while RHS contains 8 hydrogens, so adding 2 $H^{+}$ to LHS.
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) + 2 H^{+} \to 5 M n O_{2} (s) + 2 H_{2} O$

Now adding $O H^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2} O$ ,
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) + 2 H^{+} + 2 O H^{-} \to 5 M n O_{2} (s) + 4 H_{2} O + 2 O H^{-}$
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) + 2 H_{2} O \to 5 M n O_{2} (s) + 4 H_{2} O + 2 O H^{-}$

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

$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}$

This is the final balanced equation.

We can also see that charge on both sides is -2. which also indicates that the equation is balanced now.

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