8.19. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent. (a) P4(s) + OH–(aq) → PH3(g) + H2PO2–(aq) (b) N2H4(l) + ClO–(aq) → NO(g) + Cl-(aq) (c) Cl2O7(g) + H2O2(aq) → ClO2–(aq) + O2(g) + H+

(a) Oxidation number method:The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, P4? 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 H2? PO2−? with 3. P4? (s) + OH− (aq) → PH3? (g) + 3H2? PO2−? (aq)To balance O atoms, multiply OH− ions by 6.P4? (s) + 6OH− (aq) → PH3? (g) + 3H2? PO2−? (aq)To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.P4? (s) + 6OH− (aq) + 3H2? O (l) → PH3? (g) + 3H2? PO2−? (aq) + 3OH− (aq)Subtract 3 hydroxide ions from both sides.P4? (s) + 3OH− (aq) + 3H2? O (l) → PH3? (g) + 3H2? PO2−? (aq)Ion electron method:The oxidation half reaction is P4? (s) → H2? PO2−? (aq).The P atom is balanced.P4? (s) → 4H2? PO2−? (aq)The oxidation number is balanced by adding 4 electrons on RHS.P4? (s) → 4H2? PO2−? (aq) + 4e−The charge is balanced by adding 8 hydroxide ions on LHS.P4? (s)+8OH− (aq)→4H2? PO2−? (aq)The O and H atoms are balanced.The reduction half reaction is P4? (s)→PH3? (g).The oxidation number is balanced by adding 12 electrons on LHS.P4? (s)+12e−→PH3? (g)The charge is balanced by adding 12 hydroxide ions on RHS.P4? (s)+12e−→PH3? (g)+12OH−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.P4? (s)+3OH− (aq)+3H2? O (l)→PH3? (g)+3H2? PO2−? (aq) (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 CIO−3 to – 1 in Cl–. Hence, in this reaction, N2H4 is the reducing agent and CIO−3 is the oxidizing agent.Ion - electron method:The oxidation half equation is:−2 +2N2H4 (l)? NO (g)The N atoms are balanced as:N2H4 (l)?2NO (g)+8e−The charge is balanced by adding 8 OH– ions as:N2H4 (I)+8OH− (aq)?2NO (g)+8e−The O atoms are balanced by adding 6H2O as:N2H4 (I)+8OH− (aq)?2NO (g)+6H2O (I)+8e−. (i)The reduction half equation is:+5

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Redox Reactions Ncert Solutions Chemistry Class 11th Chemistry Class 11th Redox Reactions

8.19. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH^–(aq) → PH₃(g) + H₂PO₂^–(aq)
(b) N₂H₄(l) + ClO^–(aq) → NO(g) + Cl^-(aq)
(c) Cl₂O₇(g) + H₂O₂(aq) → ClO₂^–(aq) + O₂(g) + H⁺

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Vishal Baghel | Contributor-Level 10

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(a) Oxidation number method:
The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, P₄? 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?PO₂⁻? with 3.

P₄? (s) + OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance O atoms, multiply OH⁻ ions by 6.

P₄? (s) + 6OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq

...more

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying H_2?PO₂⁻? with 3.

P₄? (s) + OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance O atoms, multiply OH⁻ ions by 6.

P₄? (s) + 6OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.

P₄? (s) + 6OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq) + 3OH⁻ (aq)
Subtract 3 hydroxide ions from both sides.

P_4? (s) + 3OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

Ion electron method:
The oxidation half reaction is

P₄? (s) → H₂? PO₂⁻? (aq).
The P atom is balanced.

P₄? (s) → 4H₂? PO₂⁻? (aq)

The oxidation number is balanced by adding 4 electrons on RHS.

P₄? (s) → 4H₂? PO₂⁻? (aq) + 4e⁻

The charge is balanced by adding 8 hydroxide ions on LHS.

P₄? (s)+8OH⁻ (aq)→4H₂? PO₂⁻? (aq)
The O and H atoms are balanced.

The reduction half reaction is P₄? (s)→PH₃? (g).
The oxidation number is balanced by adding 12 electrons on LHS.

P_4? (s)+12e⁻→PH₃? (g)
The charge is balanced by adding 12 hydroxide ions on RHS.

P₄? (s)+12e⁻→PH_3? (g)+12OH⁻
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)+3OH⁻ (aq)+3H₂? O (l)→PH₃? (g)+3H₂? PO₂⁻? (aq)

(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 CIO⁻₃ to – 1 in Cl^–. Hence, in this reaction, N₂H₄ is the reducing agent and CIO⁻₃ is the oxidizing agent.
Ion - electron method:
The oxidation half equation is:

−2 +2

N₂H₄ (l)? NO (g)
The N atoms are balanced as:

N₂H₄ (l)?2NO (g)+8e⁻
The charge is balanced by adding 8 OH^– ions as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+8e⁻
The O atoms are balanced by adding 6H2O as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+6H₂O (I)+8e⁻. (i)
The reduction half equation is:

less

(a) Oxidation number method: The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence,  P4?  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 H2? PO2−? with 3. P4? (s) + OH− (aq) → PH3? (g) + 3H2? PO2−? (aq)To balance O atoms, multiply OH− ions by 6.P4? (s) + 6OH− (aq) → PH3? (g) + 3H2? PO2−? (aq)To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.P4? (s) + 6OH− (aq) + 3H2? O (l) → PH3? (g) + 3H2? PO2−? (aq) + 3OH− (aq) Subtract 3 hydroxide ions from both sides.P4? (s) + 3OH− (aq) + 3H2? O (l) → PH3? (g) + 3H2? PO2−? (aq) Ion electron method: The oxidation half reaction is P4? (s) → H2? PO2−? (aq). The P atom is balanced.P4? (s) → 4H2? PO2−? (aq)The oxidation number is balanced by adding 4 electrons on RHS.P4? (s) → 4H2? PO2−? (aq) + 4e−The charge is balanced by adding 8 hydroxide ions on LHS.P4? (s)+8OH− (aq)→4H2? PO2−? (aq) The O and H atoms are balanced.The reduction half reaction is P4? (s)→PH3? (g). The oxidation number is balanced by adding 12 electrons on LHS.P4? (s)+12e−→PH3? (g) The charge is balanced by adding 12 hydroxide ions on RHS.P4? (s)+12e−→PH3? (g)+12OH− 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.P4? (s)+3OH− (aq)+3H2? O (l)→PH3? (g)+3H2? PO2−? (aq)  (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 CIO−3 to – 1 in Cl–. Hence, in this reaction,  N2H4 is the reducing agent and CIO−3 is the oxidizing agent. Ion - electron method: The oxidation half equation is:−2               +2N2H4 (l)? NO (g) The N atoms are balanced as:N2H4 (l)?2NO (g)+8e− The charge is balanced by adding 8 OH– ions as:N2H4 (I)+8OH− (aq)?2NO (g)+8e− The O atoms are balanced by adding 6H2O as:N2H4 (I)+8OH− (aq)?2NO (g)+6H2O (I)+8e−. (i) The reduction half equation is:+5

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