Question

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

Answer 1

(a)

The O.N. (oxidation number) of P decreases from 0 in P₄ to –3 in PH₃ and increases from 0 in P₄ to + 2 in. Hence, P₄ acts both as an oxidizing agent and a reducing agent in this reaction.

Ion–electron method:

The oxidation half equation is:

${\mathrm{P}}_4\left(\mathrm{s}\right)\to {\mathrm{HPO}}_2^{-}\left(\mathrm{aq}\right)$

The P atom is balanced as:

${\mathrm{P}}_4\left(\mathrm{s}\right)\to 4{\mathrm{HPO}}_2^{-}\left(\mathrm{aq}\right)$

The O atom is balanced by adding 8 H₂O molecules:

${\mathrm{P}}_4\left(\mathrm{s}\right)+8{\mathrm{H}}_2\mathrm{O}\to 4{\mathrm{HPO}}_2^{-}\left(\mathrm{aq}\right)$

The H atom is balanced by adding 12 H⁺ ions:

${\mathrm{P}}_4\left(\mathrm{s}\right)+8{\mathrm{H}}_2\mathrm{O}\to 4{\mathrm{HPO}}_2^{-}\left(\mathrm{aq}\right)+12{\mathrm{H}}^{+}$

The charge is balanced by adding e^– as:

${\mathrm{P}}_4\left(\mathrm{s}\right)+8{\mathrm{H}}_2\mathrm{O}\to 4{\mathrm{HPO}}_2^{-}\left(\mathrm{aq}\right)+12{\mathrm{H}}^{+}+8{\mathrm{e}}^{-}$ ...(i)

The reduction half equation is:

The P atom is balanced as:

${\mathrm{P}}_4\left(\mathrm{s}\right)\to 4{\mathrm{PH}}_3\left(\mathrm{g}\right)$

The H is balanced by adding 12 H⁺ as:

${\mathrm{P}}_4\left(\mathrm{s}\right)+12{\mathrm{H}}^{+}\to 4{\mathrm{PH}}_3\left(\mathrm{g}\right)$

The charge is balanced by adding 12e^– as:

${\mathrm{P}}_4\left(\mathrm{s}\right)+12{\mathrm{H}}^{+}+12{\mathrm{e}}^{-}\to 4{\mathrm{PH}}_3\left(\mathrm{g}\right)$ ...(ii)

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:

$5{\mathrm{P}}_4\left(\mathrm{s}\right)+24{\mathrm{H}}_2\mathrm{O}\to 12{\mathrm{HPO}}_2^{-}+8{\mathrm{PH}}_3\left(\mathrm{g}\right)+12{\mathrm{H}}^{+}$

As, the medium is basic, add 12OH^– both sides as:
$5{\mathrm{P}}_4\left(\mathrm{s}\right)+12{\mathrm{H}}_2\mathrm{O}+12{\mathrm{OH}}^{-}\to 12{\mathrm{HPO}}_2^{-}+8{\mathrm{PH}}_3\left(\mathrm{g}\right)$

This is the required balanced equation.

Oxidation number method:

Let, total no of P reduced = x
$\therefore$ Total no of P oxidised = 4 – x
${\mathrm{P}}_4\left(\mathrm{s}\right)+{\mathrm{OH}}^{-}\to x{\mathrm{PH}}_3\left(\mathrm{g}\right)+\left(4-x\right){\mathrm{HPO}}_2^{-}$ ... (i)
Total decrease in oxidation number of P = x × 3 = 3x

Total increase in oxidation number of P = (4 – x) × 2 = 8 – 2x
$\because$ 3x = 8 – 2x
x = 8/5
From (i),
$5{\mathrm{P}}_4\left(\mathrm{s}\right)+5{\mathrm{OH}}^{-}\to 8{\mathrm{PH}}_3\left(\mathrm{g}\right)+12{\mathrm{HPO}}_2^{-}$
Since, reaction occures in basic medium, the charge is balanced by adding 7OH^– on LHS as:
$5{\mathrm{P}}_4\left(\mathrm{s}\right)+12{\mathrm{OH}}^{-}\to 8{\mathrm{PH}}_3\left(\mathrm{g}\right)+12{\mathrm{HPO}}_2^{-}$
The O atoms are balanced by adding 12H₂O as:
$5{\mathrm{P}}_4\left(\mathrm{s}\right)+12{\mathrm{H}}_2\mathrm{O}+12{\mathrm{OH}}^{-}\to +12{\mathrm{HPO}}_2^{-}+8{\mathrm{PH}}_3\left(\mathrm{g}\right)$
This is the required balanced equation.

(b)

The oxidation number of N increases from – 2 in N₂H₄ to + 2 in NO and the oxidation number of Cl decreases from + 5 in to – 1 in Cl^–. Hence, in this reaction, N₂H₄ is the reducing agent and is the oxidizing agent.

Ion–electron method:

The oxidation half equation is:

The N atoms are balanced as:

The oxidation number is balanced by adding 8 electrons as:

The charge is balanced by adding 8 OH^–ions as:

The O atoms are balanced by adding 6H₂O as:

The reduction half equation is:

The oxidation number is balanced by adding 6 electrons as:

The charge is balanced by adding 6OH^– ions as:

The O atoms are balanced by adding 3H₂O as:

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

Oxidation number method:

Total decrease in oxidation number of N = 2 × 4 = 8

Total increase in oxidation number of Cl = 1 × 6 = 6

On multiplying N₂H₄ with 3 and with 4 to balance the increase and decrease in O.N., we get:

The N and Cl atoms are balanced as:

The O atoms are balanced by adding 6H₂O as:

This is the required balanced equation.

(c)

The oxidation number of Cl decreases from + 7 in Cl₂O₇ to + 3 in and the oxidation number of O increases from – 1 in H₂O₂ to zero in O₂. Hence, in this reaction, Cl₂O₇ is the oxidizing agent and H₂O₂ is the reducing agent.

Ion–electron method:

The oxidation half equation is:

The oxidation number is balanced by adding 2 electrons as:

The charge is balanced by adding 2OH^–ions as:

The oxygen atoms are balanced by adding 2H₂O as:

The reduction half equation is:

The Cl atoms are balanced as:

The oxidation number is balanced by adding 8 electrons as:

The charge is balanced by adding 6OH^– as:

The oxygen atoms are balanced by adding 3H₂O as:

The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:

Oxidation number method:

Total decrease in oxidation number of Cl₂O₇ = 4 × 2 = 8

Total increase in oxidation number of H₂O₂ = 2 × 1 = 2

By multiplying H₂O₂ and O₂ with 4 to balance the increase and decrease in the oxidation number, we get:

The Cl atoms are balanced as:

The O atoms are balanced by adding 3H₂O as:

The H atoms are balanced by adding 2OH^– and 2H₂O as:

This is the required balanced equation.