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.
More Questions From Class 11 Chemistry - Chapter Redox Reactions
- Q:-
Balance the following redox reactions by ion – electron method :
(a) MnO4 – (aq) + I – (aq) → MnO2 (s) + I2(s) (in basic medium)
(b) MnO4 – (aq) + SO2 (g) → Mn2+ (aq) + HSO4– (aq) (in acidic solution)
(c) H2O2 (aq) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
- Q:-
Assign oxidation number to the underlined elements in each of the following species:
(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12 H2O
- Q:-
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) KI3
(b) H2S4O6
(c) Fe3O4
(d) CH3CH2OH
(e) CH3COOH
- Q:-
Justify that the following reactions are redox reactions:
(a) CuO(s) + H2(g) → Cu(s) + H2O(g)
(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)
(d) 2K(s) + F2(g) → 2K+F– (s)
(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
- Q:-
Fluorine reacts with ice and results in the change:
H2O(s) + F2(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
- Q:-
Write the formulae for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide - Q:-
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
- Q:-
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
(a) 2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq)
(b) HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH-(aq) → 2Ag(s) + HCOO-(aq) + 4NH3(aq) + 2H2O(l)
(c) HCHO (l) + 2Cu2+(aq) + 5 OH-(aq) → Cu2O(s) + HCOO-(aq) + 3H2O(l)
(d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)
(e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
- Q:-
The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
- Q:-
Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2- 7 and NO– 3. Suggest structure of these compounds. Count for the fallacy.
Popular Questions of Class 11 Chemistry
- Q:-
How do you account for the formation of ethane during chlorination of methane?
- Q:-
What are hybridisation states of each carbon atom in the following compounds ?
(i) CH2=C=O,
(ii) CH3CH=CH2,
(iii) (CH3)2CO,
(iv) CH2=CHCN,
(v) C6H6
- Q:-
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
- Q:-
What are the common physical and chemical features of alkali metals?
- Q:-
Calculate the molecular mass of the following:
(i) H2O
(ii) CO2
(iii) CH4 - Q:-
What is the basic theme of organisation in the periodic table?
- Q:-
Explain the formation of a chemical bond.
- Q:-
Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only. - Q:-
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure? - Q:-
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Recently Viewed Questions of Class 11 Chemistry
- Q:-
What do you understand by the term “non-stoichiometric hydrides”? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.
- Q:-
Write the favourable factors for the formation of ionic bond.
- Q:-
How would you explain the lower atomic radius of Ga as compared to Al?
- Q:-
For the reaction
2 A(g) + B(g) → 2D(g)
ΔU0 = –10.5 kJ and ΔS0 = –44.1 JK–1.
Calculate ΔG0 for the reaction, and predict whether the reaction may occur spontaneously.
- Q:-
Why is benzene extra ordinarily stable though it contains three doublebonds?
- Q:-
In the modern periodic table, the period indicates the value of:
(a) Atomic number
(b) Atomic mass
(c) Principal quantum number
(d) Azimuthal quantum number. - Q:-
Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction methods?
- Q:-
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
- Q:-
Explain, giving reasons, which of the following sets of quantum numbers are not possible.
a n = 0 l = 0 ml = 0
b n = 1 l = 0 ml = 0
c n = 1 l = 1 ml = 0
d n = 2 l = 1 ml = 0
e n = 3 l = 3 ml = – 3
f n = 3 l = 1 ml = 0
- Q:-
Explain what happens when boric acid is heated.
1 Comment(s) on this Question
Thanks for our information
- All Chapters Of Class 11 Chemistry
- All Subjects Of Class 11