Question

Balance the following equation by ion electron method.
$Mn{O}_4^{2-}+H_{2}O\to Mn{O}_4^{-}+O{H}^{-}+Mn{O}_2$.

Answer 1

Write the reaction in two half reactions,oxidation half and reduction half as:

Oxidation:$Mn{O}_4^{2-}\to Mn(O{)}_4^{-}$

Reduction:${MnO}_4^{2-}\to Mn{O}_2$

Now balance the oxygen atoms by adding $H_{2}O$ molecules.

$Mn{O}_4^{2-}\to Mn(O{)}_4^{-}$

${MnO}_4^{2-}\to Mn{O}_2+2H_{2}O$

Now balance hydrogen atoms by adding $H^{+}$ ions.

$Mn{O}_4^{2-}\to Mn(O{)}_4^{-}$

${MnO}_4^{2-}+4H^{+}\to Mn{O}_2+2H_{2}O$

As reaction takes place in basic medium, add one $O{H}^{-}$ each side for every $H^{+}$ ion present in the reaction.Combine $O{H}^{-}$ and $H^{+}$ to form $H_{2}O$.

$Mn{O}_4^{2-}\to Mn(O{)}_4^{-}$

${MnO}_4^{2-}+2H_{2}O\to Mn{O}_2+4O{H}^{-}$

To balance the charge,add electrons to more positive side to equal the less positive side of the half reaction.

$Mn{O}_4^{2-}\to Mn(O{)}_4^{-}+e^{-}$

${MnO}_4^{2-}+2H_{2}O+2e^{-}\to Mn{O}_2+4O{H}^{-}$

Now multiply oxidation half reaction by 2 and add both the reactions we will get

$3{MnO}_4^{2-}+2H_{2}O\to 2Mn(O{)}_4^{-}+4{OH}^{-}+Mn{O}_2$