Name: Hydrogen Electrode
Uploaded: 2019-07-19
Duration: 3 min 7 s
Description: Every unit of measurement has a standard measurement associated with it. The same is with the electrode potential. Let's learn what the standard measurement...

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,
$3 {\overset{- 2}{N}}_{2} H_{4} (g) + 4 \overset{+ 5}{Cl} O_{3}^{-} (aq) \to 6 \overset{+ 2}{N} O (g) + 4 {Cl}^{-} (g) + 6 H_{2} O (l)$

NCERT Solutions for Chapter: Redox Reactions, Exercise 1: EXERCISES

NCERT Chemistry Solutions for Exercise - NCERT Solutions for Chapter: Redox Reactions, Exercise 1: EXERCISES

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