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)

Step 1: The two half reactions involved in the given reaction are: -1 0 Oxidation half reaction: l (aq) → l2(s) +7 +4 Reduction half reaction: Mn O-4(aq) → MnO2(aq) Step 2: Balancing I in the oxidation half reaction, we have: 2l-(aq) → l2(s) Now, to balance the charge, we add 2 e- to the RHS of the reaction. 2l-(aq) → l2(s) + 2e- Step 3 : In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction. MnO-4(aq) + 3e- →MnO2(aq) Now, to balance the charge, we add 4 OH- ions to the RHS of the reaction as the reaction is taking place in a basic medium. MnO-4(aq) + 3e- →MnO2(aq) + 4OH- Step 4: In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS. MnO-4(aq) + 2H2O + 3e- →MnO2(aq) + 4OH- Step 5: Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have: 6l-(aq) → 3l2(s) + 2e- 2MnO-4(aq) + 4H2O + 6e- → 2MnO2(s) + 8OH-(aq) Step 6: Adding the two half reactions, we have the net balanced redox reaction as: 6l-(aq) + 2MnO-4(aq) + 4H2O(l) → 3l2(s) + 2MnO2(s) + 8OH-(aq) (b) Following the steps as in part (a), we have the oxidation half reaction as: SO2(g) + 2H2O(l) → HSO-4(aq) + 3H+(aq) + 2e-(aq) And the reduction half reaction as: MnO-4(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l) Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as: 2MnO-4(aq) + 5SO2(g) + 2H2O(l) + H+(aq) → Mn2+(aq) + HSO-4(aq) (c) Following the steps as in part (a), we have the oxidation half reaction as: Fe2+(aq) → Fe3+(aq) + e- And the reduction half reaction as: H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l) Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as: H2O2(aq) + 2Fe2+(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l) (d) Following the steps as in part (a), we have the oxidation half reaction as: SO2(g) + 2H2O(l) → SO2-4(aq) + 4H+ (aq) + 2e- And the reduction half reaction as: Cr2O2-7(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 3SO2-4(aq) + H2O(l) Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as: Cr2O2-7(aq) + 3SO2(g) + 2H+(aq) → 2Cr3+(aq) + 3SO2-4(aq) + H2O(l)

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Balance the following redox reactions by | Class 11 Chemistry Chapter Redox Reactions, Redox Reactions NCERT Solutions

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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 18:

Balance the following redox reactions by ion – electron method :

(a) MnO₄^–(aq) + I^–(aq) → MnO_{2 (s)} + I_2(s) (in basic medium)

(b) MnO₄ ^– (aq) + SO_{2 (g)}→ Mn²⁺ (aq) + HSO₄^– (aq) (in acidic solution)

(d) Cr₂O₇ ^2– + SO_2(g) → Cr³⁺ _(aq) + SO₄^2–(aq) (in acidic solution)

. 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.

Question 18:

Balance the following redox reactions by ion – electron method :

(a) MnO₄^–(aq) + I^–(aq) → MnO_{2 (s)} + I_2(s) (in basic medium)

(b) MnO₄ ^– (aq) + SO_{2 (g)}→ Mn²⁺ (aq) + HSO₄^– (aq) (in acidic solution)

(d) Cr₂O₇ ^2– + SO_2(g) → Cr³⁺ _(aq) + SO₄^2–(aq) (in acidic solution)

Answer:

Step 1:

The two half reactions involved in the given reaction are:

-1 0

Oxidation half reaction: l_(aq) → l_2(s)

+7 +4

Reduction half reaction: Mn O^-_4(aq) → MnO_2(aq)

Step 2:

Balancing I in the oxidation half reaction, we have:

2l^-_(aq)→ l_2(s)

Now, to balance the charge, we add 2 e^- to the RHS of the reaction.

2l^-_(aq)→ l_2(s)+ 2e^-

Step 3 :

In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.

MnO^-_4(aq) + 3e^- →MnO_2(aq)

Now, to balance the charge, we add 4 OH^-ions to the RHS of the reaction as the reaction is taking place in a basic medium.

MnO^-_4(aq) + 3e^- →MnO_2(aq)+ 4OH^-

Step 4:

In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.

MnO^-_4(aq) + 2H₂O + 3e^- →MnO_2(aq)+ 4OH^-

Step 5:

Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:

6l^-_(aq) → 3l_2(s)+ 2e^-

2MnO^-_4(aq) + 4H₂O + 6e^- → 2MnO_2(s)+ 8OH^-_(aq)

Step 6:

Adding the two half reactions, we have the net balanced redox reaction as:

6l^-_(aq)+ 2MnO^-_4(aq) + 4H₂O_(l) → 3l2(s) + 2MnO_2(s)+ 8OH^-_(aq)

(b) Following the steps as in part (a), we have the oxidation half reaction as:

SO_2(g) + 2H₂O_(l) → HSO^-_4(aq)+ 3H⁺_(aq) + 2e^-_(aq)

And the reduction half reaction as:

MnO^-_4(aq) + 8H⁺_(aq) + 5e^- → Mn²⁺_(aq) + 4H₂O_(l)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:

2MnO^-_4(aq) + 5SO_2(g)+ 2H₂O_(l) + H⁺_(aq) → Mn²⁺_(aq) + HSO^-_4(aq)

Fe²⁺_(aq) → Fe³⁺_(aq)+ e^-

And the reduction half reaction as:

H₂O_2(aq) + 2H⁺_(aq) + 2e^- → 2H₂O_(l)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

H₂O_2(aq) + 2Fe²⁺_(aq) + 2H⁺_(aq) → 2Fe³⁺_(aq)+ 2H₂O_(l)

(d) Following the steps as in part (a), we have the oxidation half reaction as:

SO_2(g)+ 2H₂O_(l) → SO^2-_4(aq) + 4H⁺ _(aq) + 2e^-

And the reduction half reaction as:

Cr₂O^2-_7(aq) + 14H⁺_(aq) + 6e^-→ 2Cr³⁺_(aq)+ 3SO^2-_4(aq) + H₂O_(l)

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

Cr₂O^2-_7(aq) + 3SO_2(g)+ 2H⁺_(aq) → 2Cr³⁺_(aq)+ 3SO^2-_4(aq)+ H₂O_(l)

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Other Questions of Chapter Redox Reactions

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?

Consider the reactions:

(a) H₃PO_2(aq) + 4 AgNO_3(aq) + 2 H₂O_(l) → H₃PO_4(aq) + 4Ag_(s)+ 4HNO_3(aq)

(b) H₃PO_2(aq) + 2CuSO_4(aq) + 2 H₂O_(l) → H₃PO_4(aq) + 2Cu_(s) + H₂SO_4(aq)

(d) C₆H₅CHO_(l) + 2Cu²⁺_(aq) + 5OH^-_(aq)→ No change observed.

What inference do you draw about the behaviour of Ag⁺and Cu²⁺from these reactions?

Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O^2- ₇ and NO^– ₃. Suggest structure of these compounds. Count for the fallacy.

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

The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO², and H⁺ ion. Write a balanced ionic equation for the reaction.

Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non metals that can show disproportionation reaction.

(b) Select three metals that can show disproportionation reaction.

Consider the elements:

Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only postive oxidation state.

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO₃ with silver electrodes

(ii) An aqueous solution AgNO₃ with platinum electrodes

(iii) A dilute solution of H₂SO₄ with platinum electrodes

(iv) An aqueous solution of CuCl₂ with platinum electrodes.

Consider the reactions:

(a) 6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(aq) + 6O_2(g)

(b) O_3(g)+ H₂O_2(l) → H₂O_(l)+ 2O_2(g)

Why it is more appropriate to write these reactions as:

(a) 6CO_2(g) + 12H₂O_(l) → C₆H₁₂O_6(aq) + 6H₂O_(l)+ 6O_2(g)

(b) O_3(g) + H₂O_2(l) → H₂O_(l)+ O_2(g) + O_2(g)

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Popular questions of class 11 Chemistry

Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

The mass of an electron is 9.1 × 10^–31 kg. If its K.E. is 3.0 × 10^–25 J, calculate its wavelength.

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(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Which one of the following will have largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl2(g)

Density of a gas is found to be 5.46 g/dm³ at 27 °C at 2 bar pressure. What will be its density at STP?

What is meant by the term bond order? Calculate the bond order of: N₂, O₂,O₂⁺,and O₂^-.

Recently Viewed Questions of Class 11 Chemistry

How would you convert the following compounds into benzene?

(i) Ethyne (ii) Ethene (iii) Hexane

Explain structures of diborane and boric acid.

For the following equilibrium, K_c= 6.3 × 10¹⁴ at 1000 K

NO (g) + O₃ (g) ↔ NO₂ (g) + O₂(g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?

Describe the change in hybridisation (if any) of the Al atom in the following reaction.

Find out the value of K_c for each of the following equilibria from the value of K_p:

(i) 2NOCl (g) ↔ 2NO (g) + Cl₂ (g); Kp = 1.8 × 10^–2 at 500 K

(ii) CaCO₃ (s) ↔ CaO(s) + CO₂(g); Kp = 167 at 1073 K

In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

Find energy of each of the photons which

(i) correspond to light of frequency 3× 1015 Hz.

(ii) have wavelength of 0.50 Å.

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^–1 K^–1.

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?

Write Lewis symbols for the following atoms and ions:

S and S2–; Al and Al3+; H and H–

Study Tips for Answering NCERT Questions:

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:

Read the question carefully and focus on the core concept being asked.
Reference examples and data from the chapter when answering questions about Redox Reactions.
Review previous year question papers to get an idea of how such questions may be framed in exams.
Practice answering questions within the time limit to improve your speed and accuracy.
Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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