Class 11 Chemistry - Chapter Redox Reactions NCERT Solutions | Balance the following equations in basic
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.
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Balance the following redox reactions by ion – electron method :
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