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

$P_{4} (s) + O H^{-} (a q) \to P H_{3} (g) + H_{2} P O_{2}^{-} (a q)$

Open in App

Solution

Oxidation number method:
The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, $P_{4}$ is oxidizing as well as reducing agent.
During reduction, the total decrease in the oxidation number for 4 P atoms is 12.
During oxidation, total increase in the oxidation number for 4 P atoms is 4.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $H_{2} P O_{2}^{-}$ with 3.
$P_{4} (s) + O H^{-} (a q) \to P H_{3} (g) + 3 H_{2} P O_{2}^{-} (a q)$
To balance O atoms, multiply $O H^{-}$ ions by 6.
$P_{4} (s) + 6 O H^{-} (a q) \to P H_{3} (g) + 3 H_{2} P O_{2}^{-} (a q)$
To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.
$P_{4} (s) + 6 O H^{-} (a q) + 3 H_{2} O (l) \to P H_{3} (g) + 3 H_{2} P O_{2}^{-} (a q) + 3 O H^{-} (a q)$
Subtract 3 hydroxide ions from both sides.
$P_{4} (s) + 3 O H^{-} (a q) + 3 H_{2} O (l) \to P H_{3} (g) + 3 H_{2} P O_{2}^{-} (a q)$
Ion electron method:
The oxidation half reaction is $P_{4} (s) \to H_{2} P O_{2}^{-} (a q)$ .
The P atom is balanced.
$P_{4} (s) \to 4 H_{2} P O_{2}^{-} (a q)$
The oxidation number is balanced by adding 4 electrons on RHS.
$P_{4} (s) \to 4 H_{2} P O_{2}^{-} (a q) + 4 e^{-}$
The charge is balanced by adding 8 hydroxide ions on LHS.
$P_{4} (s) + 8 O H^{-} (a q) \to 4 H_{2} P O_{2}^{-} (a q)$
The O and H atoms are balanced.
The reduction half reaction is $P_{4} (s) \to P H_{3} (g)$ .
The oxidation number is balanced by adding 12 eelctrons on LHS.
$P_{4} (s) + 12 e^{-} \to P H_{3} (g)$
The charge is balanced by adding 12 hydroxide ions on RHS.
$P_{4} (s) + 12 e^{-} \to P H_{3} (g) + 12 O H^{-}$
The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2.
The half reactions are then added to obtain balanced chemical equation.
$P_{4} (s) + 3 O H^{-} (a q) + 3 H_{2} O (l) \to P H_{3} (g) + 3 H_{2} P O_{2}^{-} (a q)$

Suggest Corrections

Similar questions

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
$P_{4 (s)} + O H -_{(a q)} ⟶ P H_{3 (g)} + H P O_{2 (a q)}^{-}$

$N_{2} H_{4 (I)} + C I O_{3 (a q)}^{-} ⟶ N O_{(g)} + C I_{(g)}^{-}$

$C l_{2} O_{7 (g)} + H_{2} O_{2 (a q)} ⟶ C I O_{2 (a q)}^{-} + O_{2 (g)} + H_{(a q)}^{+}$

Q. Balance the following equation by ion electron method.

P_{4} + O H^{-} + H_{2} O \to H_{2} P O_{2}^{-} + P H_{3}

Q. Balance the following equations in acidic medium by both oxidation number and ion electron methods and identify the oxidants and the reductants :

M n O_{4}^{-} (a q) + C_{2} H_{2} O_{4} (a q) \to M n^{2 +} (a q) + C O_{2} (g) + H_{2} O (l)

Q. Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction.

{H C H O}_{(l)} + 2 [A g (N H_{3})_{2}]_{(a q)}^{+} + 3 O H_{(a q)}^{-} \to 2 A g_{(s)} + H C O O_{(a q)}^{-} + 4 {N H_{3}}_{(a q)} + 2 {H_{2} O}_{(l)}

Q. Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction-

2 A g B r (s) + C_{6} H_{6} O_{2} (a q) \to 2 A g (s) + 2 H B r (a q) + C_{6} H_{4} O_{2} (a q)

Join BYJU'S Learning Program

Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent. P4(s)+OH−(aq)→PH3(g)+H2PO−2(aq)

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

$P_{4} (s) + O H^{-} (a q) \to P H_{3} (g) + H_{2} P O_{2}^{-} (a q)$