Question

Starting with the correctly balanced half-reactions, write the overall net ionic reaction for the following change:

Nitrous acid $(H N O_{2})$ reduces ${MnO}_{4}^{-}$ ( in acid solution). Also, find out the change in the oxidation number of the underlined atom.

Accepted Answer

For balancing the given reaction,

Step 1. First, write the skeletal equation.

${MnO}_{4}^{-} + {HNO}_{2} + H^{+} \to {HNO}_{3} + H_{2} O + {Mn}^{2 +}$

Step.2 Write the oxidation number of all the elements above their respective symbols.

$\overset{+ 7}{Mn} {\overset{- 2}{O}}_{4}^{-} + \overset{+ 1}{H} \overset{+ 3}{N} {\overset{- 2}{O}}_{2} + \overset{+ 1}{H^{+}} \to \overset{+ 1}{H} \overset{+ 5}{N} {\overset{- 2}{O}}_{3} + {\overset{+ 1}{H}}_{2} \overset{- 2}{O} \overset{2 +}{+ {Mn}^{2 +}}$

Step.3 Find out the species which have been oxidised and reduced and split the given skeletal equation into two half-reactions.

Since the oxidation number of $Mn$ decreases from $+ 7$ in ${MnO}_{4}^{-}$ to $+ 2$ in ${Mn}^{2 +}$ and that of $N$ increases from $+ 3$ in ${HNO}_{2}$ to $+ 5$ in ${HNO}_{3}$ . Thus, the above equation can be divided into two half-reactions.

Oxidation half equation: ${NO}_{2}^{-} \to {NO}_{3}^{-} \dots \dots \dots \dots \dots \dots \dots (i)$ (Here, $H^{+}$ are neglected from both sides for simplicity)

Reduction half equation : ${MnO}_{4}^{-} \to {Mn}^{2 +} \dots \dots \dots \dots \dots \dots (ii)$

Step.4 Balance oxidation and reduction half-reactions both.

$(a)$ Balance all the atoms other than $O$ and $H$ .

$N$ and $Mn$ are already balanced.

$(b)$ Balance $O$ atoms by adding $H_{2} O$ .

${NO}_{2}^{-} + H_{2} O \to {NO}_{3}^{-} \dots \dots \dots \dots \dots \dots (iii)$

${MnO}_{4}^{-} \to {Mn}^{2 +} + 4 H_{2} O \dots \dots \dots \dots \dots \dots (iv)$

$(c)$ Balance $H$ atoms by adding $H^{+}$ .

${NO}_{2}^{-} + H_{2} O \to {NO}_{3}^{-} + 2 H^{+} \dots \dots \dots \dots (v)$

${MnO}_{4}^{-} + 8 H^{+} \to {Mn}^{2 +} + 4 H_{2} O \dots \dots \dots \dots \dots (vi)$

$(d)$ Now, balance the charge of both sides.

${NO}_{2}^{-} + H_{2} O \to {NO}_{3}^{-} + 2 H^{+} + 2 e^{-} \dots \dots \dots \dots \dots (vii)$

${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} + 4 H_{2} O \dots \dots \dots \dots \dots (viii)$

Step.5 To balance the electrons lost in the equation $(vii)$ and gained in equation $(viii)$ , multiply eq. $(vii)$ by $5$ and eq. $(viii)$ by $2$ and add both equations.

$5 {NO}_{2}^{-} + 5 H_{2} O \to 5 {NO}_{3}^{-} + 10 H^{+} + 10 e^{-}$

$2 {MnO}_{4}^{-} + 16 H^{+} + 10 e^{-} \to 2 {Mn}^{2 +} + 8 H_{2} O$

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$2 {MnO}_{4}^{-} + 6 H^{+} + 5 {NO}_{2}^{-} \to 5 {NO}_{3}^{-} + 3 H_{2} O + 2 {Mn}^{2 +}$

And the oxidation number of $N$ from $+ 3$ in ${HNO}_{2}$ to $+ 5$ in ${HNO}_{3}$ .

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