Name: Balancing Redox Reactions with Oxidation Numbers Method
Uploaded: 2019-07-19
Duration: 8 min 51 s
Description: The oxidation number is the total number of charges on an ion formed by the gain or loss of electrons. How do we balance redox reactions by using them? Learn...

Question

Balance the following equation in the basic medium by ion-electron method and oxidation number methods and identify the oxidizing agent and the reducing agent.

N_{2} H_{4} (l) + {ClO}_{3}^{-} (aq) \to NO (g) + {Cl}^{-} (g)

Accepted Answer

$N_{2} H_{4} (l) + {ClO}_{3}^{-} (aq) \to NO (g) + {Cl}^{-} (g)$

The oxidation number (O.N.) of $Cl$ decreases from $+ 5$ in ${ClO}_{3}^{-}$ to $- 1$ in ${Cl}^{-}$ and the O.N. of nitrogen increases from $- 2$ in $N_{2} H_{4}$ to $+ 2$ in $NO$ . Hence, ${ClO}_{3}^{-}$ acts as an oxidizing agent and $N_{2} H_{4}$ acts as a reducing agent in this reaction.

Ion–electron method:
The oxidation half equation is:
$\overset{- 2}{N_{2}} \overset{+ 1}{H_{4}} (l) \to \overset{+ 2}{N} \overset{- 2}{O} (g)$
The $N$ atom is balanced as:
$N_{2} H_{4} (l) \to 2 NO (g)$

Balance oxidation number by adding electrons:
$N_{2} H_{4} (l) \to 2 NO (g) + 8 e^{-}$

As, the medium is basic, charge is balanced by $O H^{-}$ ions:
$N_{2} H_{4} (l) + 8 {OH}^{-} (aq) \to 2 NO (g) + 8 e^{-}$

The $O$ atom is balanced by adding $6 H_{2} O$ molecules:
$N_{2} H_{4} (l) + 8 {OH}^{-} (aq) \to 2 NO (g) + 8 e^{-} + 6 H_{2} O (l)$ $. . . . . . (i)$

The reduction half equation is:
$\overset{+ 5}{Cl} \overset{- 2}{O_{3}^{-}} (aq) \to \overset{- 1}{{Cl}^{-}} (g)$

The number of $Cl$ atom is already balanced. So, balance the oxidation number by adding electrons.
${ClO}_{3}^{-} (aq) + 6 e^{-} \to {Cl}^{-} (g)$

As the medium is basic the charge is balanced by ${OH}^{-}$
${ClO}_{3}^{-} (aq) + 6 e^{-} \to {Cl}^{-} (g) + 6 {OH}^{-} (aq)$

The $O$ atom is balanced by adding $H_{2} O$ molecules:
${ClO}_{3}^{-} (aq) + 6 e^{-} + 3 H_{2} O (l) \to {Cl}^{-} (g) + 6 {OH}^{-} (aq)$ $. . . . . . (ii)$

To get the final required balanced equation, multiply eqn. $(i)$ by $3$ and $(ii)$ by $4$ and add,

$N_{2} H_{4} (l) + 8 {OH}^{-} (aq) \to 2 NO (g) + 8 e^{-} + 6 H_{2} O (l) \times 3 + {ClO}_{3}^{-} (aq) + 6 e^{-} + 3 H_{2} O (l) \to {Cl}^{-} (g) + 6 {OH}^{-} (aq) \times 4$
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$3 N_{2} H_{4} (l) + 4 {ClO}_{3}^{-} (aq) \to 6 NO (g) + 4 {Cl}^{-} (g) + 6 H_{2} O (l)$

Oxidation no. method:
Step 1:
We assign oxidation nos. and balance the increase and decrease,
$3 {\overset{- 2}{N}}_{2} H_{4} (g) + 4 \overset{+ 5}{Cl} O_{3}^{-} (aq) \to 6 \overset{+ 2}{N} O (g) + 4 {Cl}^{-} (g)$

Step 2: Balance charges by adding ${OH}^{-}$ , add water molecules to balance the hydrogens,
$3 {\overset{- 2}{N}}_{2} H_{4} (g) + 4 \overset{+ 5}{Cl} O_{3}^{-} (aq) \to 6 \overset{+ 2}{N} O (g) + 4 {Cl}^{-} (g) + 6 H_{2} O (l)$

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