• Home
  • NCERT Solutions
  • Class 11
  • Chemistry
  • Redox Reactions
  • 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)

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

Q18.

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)

 

👍 0
👎 0
✍️ Add Answer
🚩 Report

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.

Important Questions & Answers

Why is this answer important for exams?

This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.

Which NCERT concept is used in this question?

This question is based on core NCERT concepts explained in the chapter and should be revised thoroughly before exams.

What common mistakes should be avoided in this question?

Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

What is the correct answer to: 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:

                        &nb...

How do you solve 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 by step?

Step-by-step explanation:
• Step 1:

• The two half reactions involved in the given reaction are:

•                                            -1              0

Latest Blog Posts

Stay updated with our latest educational content and study tips

Simple and Compound Interest Formulas with Questions

Simple and Compound Interest Formulas with Questions

Interest is one of the most significant ideas in maths and financial calculations. It is very common in banking, loan applications, investments, saving account and competitive exams. Simple Interest and Compound Interest make it easier to find out how much more is earned and/or paid on a principal during the period of time. These Interest … Read more

Read More
Spoken English Course Topics for Beginners in 2026

Spoken English Course Topics for Beginners in 2026

In 2026, students, professionals, and job seekers must have a high level of spoken English. Spoken English is a beginner level course to enhance learners’ communication skills in English, their pronunciation, vocabulary, grammar structure and confidence in speaking English fluently. Today, the emphasis of spoken English courses is placed on actual conversations, everyday speech practice … Read more

Read More
Difference Between AI and Machine Learning

Difference Between AI and Machine Learning

Artificial Intelligence (AI) and Machine Learning (ML) are two of the most sought-after technologies in today’s digital era. Although these terms are often used together, they are not the same. Machine Learning is, in fact, a component of Artificial Intelligence that enables systems to learn and enhance on their own, without direct programming. AI is … Read more

Read More
Time, Speed and Distance Formulas

Time, Speed and Distance Formulas

Time, Speed, and Distance are some of the most important concepts in mathematics and aptitude. These are the concepts which are used to calculate the speed of an object, time taken for an object to move and the distance traveled during the motion. Questions from this topic are frequently seen in School Tests, Competitive Tests, … Read more

Read More

Student Discussion

Be the first to comment.

ADD NEW COMMENT

        Can’t find your school? Type full name and submit.