Academic session 2026-27, Chapter 7 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Redox Reactions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 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) + HPO2 – (aq)
(b) N2H4(l) + ClO3 – (aq) → NO(g) + Cl–(g)
(c) Cl2O7 (g) + H2O2(aq) → ClO – 2(aq) + O2(g) + H + (aq)
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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) + HPO2 – (aq)
(b) N2H4(l) + ClO3 – (aq) → NO(g) + Cl–(g)
(c) Cl2O7 (g) + H2O2(aq) → ClO – 2(aq) + O2(g) + H + (aq)
(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 HPO-2. Hence, P4 acts both as an oxidizing agent as well as a reducing agent in this reaction.
Ion-electron method:
The oxidation half equation is:
P4(s) → H2PO-(aq)
The P atom is balanced as:
P0 4(s) → 4H2P+1O-(aq)
The O.N. is balanced by adding 4 electrons as:
P4(s) → 4H2PO-(aq) + 4e-
The charge is balanced by adding 8OH- as:
P4(s) + 8OH - (aq) → 4H2PO-2(aq)
The O and H atoms are already balanced. The reduction half equation is:
P4(s) → PH3(g)
The P atom is balanced as
P04(s) → 4 P-3H3(g)
The O.N. is balanced by adding 12 electrons as:
P4(s) + 12e- → 4 PH3(g)
The charge is balanced by adding 12OH- as:
P4(s) + 12e- → 4 PH3(g) + 12OH-(aq) .....(i)
The O and H atoms are balanced by adding 12H2O as:
P4(s) + 12H2O(l) + 12e- → 4 PH3(g) + 12OH-(aq) -- (ii)
By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:
P4(s) + 3OH-(aq) + 3H2O → PH3 + 3H2PO-2(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:
N-22 H4(l) → N+2 O(g)
The N atoms are balanced as:
N2H4(l) → 2NO(g)
The oxidation number is balanced by adding 8 electrons as:
N2H4(l) → 2NO(g) + 8e-
The charge is balanced by adding 8 OH-ions as:
N2H4(l) + 8OH-(aq) → 2NO(g) + 8e-
The O atoms are balanced by adding 6H2O as:
N2H4(l) + 8OH-(aq) → 2NO(g) + 6H2O(l) + 8e- .... (i)
The reduction half equation is:
C+5IO-3(aq) → C-1l-(aq)
The oxidation number is balanced by adding 6 electrons as:
CIO-3(aq) + 6e- → Cl-(aq)
The charge is balanced by adding 6OH- ions as:
CIO-3(aq) + 6e- → Cl-(aq) + 6OH-(aq)
The O atoms are balanced by adding 3H2O as:
CIO-3(aq) + 3H2O(l) + 6e- → Cl-(aq) + 6OH-(aq) .... (ii)
The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:
3N2H4(l) + 4CIO-3(aq) → 6NO(g) + 4Cl-(aq) + 6H2O(l)
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 CIO-3 with 4 to balance the increase and decrease in O.N., we get:
3N2H4(l) + 4CIO-3(aq) → NO(g) + Cl-(aq)
The N and Cl atoms are balanced as:
3N2H4(l) + 4CIO-3(aq) → 6NO(g) + 4Cl-(aq)
The O atoms are balanced by adding 6H2O as:
3N2H4(l) + 4CIO-3(aq) → 6NO(g) + 4Cl-(aq) + 6H2O(l)
This is the required balanced equation.
(c)
The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in CIO-2and 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:
H2O-12(aq) → O02(g)
The oxidation number is balanced by adding 2 electrons as:
H2O2(aq) → O2(g) + 2e-
The charge is balanced by adding 2OH-ions as:
H2O2(aq) + 2OH-(aq) → O2(g) + 2e-
The oxygen atoms are balanced by adding 2H2O as:
H2O2(aq) + 2OH-(aq) → O2(g) + 2H2O(l) + 2e- ... (i)
The reduction half equation is:
C+7l2O7(g) → C+3lO-2(g)
The Cl atoms are balanced as:
Cl2O7(g) → 2ClO-2(g)
The oxidation number is balanced by adding 8 electrons as:
Cl2O7(g) + 8e- → 2ClO-2(g)
The charge is balanced by adding 6OH- as:
Cl2O7(g) + 8e- → 2ClO-2(g) + 6OH- (aq)
The oxygen atoms are balanced by adding 3H2O as:
Cl2O7(g) + 3H2O(l) + 8e- → 2ClO-2(g) + 6OH- (aq) .... (ii)
The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:
Cl2O7(g) + 4H2O2(aq) + 2OH- (aq) → 2ClO-2(aq) + 4O2(g) + 5H2O(l)
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:
Cl2O7(g) + 4H2O2(aq) → CIO-2(aq) + 4O2(g)
The Cl atoms are balanced as:
Cl2O7(g) + 4H2O2(aq) → 2CIO-2(aq) + 4O2(g)
The O atoms are balanced by adding 3H2O as:
Cl2O7(g) + 4H2O2(aq) → 2CIO-2(aq) + 4O2(g) + 3H2O(l)
The H atoms are balanced by adding 2OH- and 2H2O as:
Cl2O7(g) + 4H2O2(aq) + 2OH-(aq) → 2CIO-2(aq) + 4O2(g) + 5H2O(l)
This is the required balanced equation.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 19: Balance the following equations in basic medium by ion-electron method and oxidation number methods ....
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