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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)?
(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)?
Most Upvoted Answer
Balance the following redox reactions by ion – electron method :(a) Mn...
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)
Community Answer
Balance the following redox reactions by ion – electron method :(a) Mn...
Redox reactions can be balanced by the ion-electron method. The steps involved are:
1. Write the unbalanced equation for the reaction.
2. Separate the equation into half-reactions, one for oxidation and one for reduction.
3. Balance the atoms and charges in each half-reaction.
4. Multiply each half-reaction by a coefficient so that the number of electrons gained in reduction equals the number of electrons lost in oxidation.
5. Add the half-reactions together and cancel out any common terms.
6. Check that the charge and atom balance is correct.
(a) MnO4 – (aq) I – (aq) → MnO2 (s) I2(s) (in basic medium)
- Separate the equation into half-reactions:
MnO4 – → MnO2
I – → I2
- Balance the atoms and charges in each half-reaction:
MnO4 – → MnO2 + 2H2O + 3e–
I – + 2e– → I2
- Multiply each half-reaction by a coefficient so that the number of electrons gained in reduction equals the number of electrons lost in oxidation:
3I – + 6OH– → I3– + 3H2O + 6e–
2MnO4 – + 6e– + 4H2O → 2MnO2 + 8OH–
- Add the half-reactions together and cancel out any common terms:
3I – + 2MnO4 – + 6OH– → I3– + 2MnO2 + 4H2O
- Check that the charge and atom balance is correct.
(b) MnO4 – (aq) SO2 (g) → Mn2 (aq) HSO4– (aq) (in acidic solution)
- Separate the equation into half-reactions:
MnO4 – → Mn2+
SO2 → HSO4–
- Balance the atoms and charges in each half-reaction:
MnO4 – + 8H+ + 5e– → Mn2+ + 4H2O
SO2 + 2H2O + 2e– → HSO4– + 2H+
- Multiply each half-reaction by a coefficient so that the number of electrons gained in reduction equals the number of electrons lost in oxidation:
5SO2 + 2MnO4 – + 16H+ → 2Mn2+ + 5HSO4– + 8H2O
- Add the half-reactions together and cancel out any common terms:
5SO2 + 2MnO4 – + 16H+ → 2Mn2+ + 5HSO4– + 8H2O
- Check that the charge and atom balance is correct.
(c) H2O2 (aq) Fe2+ (aq) → Fe3+ (aq) H2O (l) (in acidic solution)
- Separate the equation into half-reactions:
H2O2 → H2O
Fe2+ → Fe3+
- Balance the atoms and charges in each half-reaction:
H2O2 + 2H+ + 2e– →
1. Write the unbalanced equation for the reaction.
2. Separate the equation into half-reactions, one for oxidation and one for reduction.
3. Balance the atoms and charges in each half-reaction.
4. Multiply each half-reaction by a coefficient so that the number of electrons gained in reduction equals the number of electrons lost in oxidation.
5. Add the half-reactions together and cancel out any common terms.
6. Check that the charge and atom balance is correct.
(a) MnO4 – (aq) I – (aq) → MnO2 (s) I2(s) (in basic medium)
- Separate the equation into half-reactions:
MnO4 – → MnO2
I – → I2
- Balance the atoms and charges in each half-reaction:
MnO4 – → MnO2 + 2H2O + 3e–
I – + 2e– → I2
- Multiply each half-reaction by a coefficient so that the number of electrons gained in reduction equals the number of electrons lost in oxidation:
3I – + 6OH– → I3– + 3H2O + 6e–
2MnO4 – + 6e– + 4H2O → 2MnO2 + 8OH–
- Add the half-reactions together and cancel out any common terms:
3I – + 2MnO4 – + 6OH– → I3– + 2MnO2 + 4H2O
- Check that the charge and atom balance is correct.
(b) MnO4 – (aq) SO2 (g) → Mn2 (aq) HSO4– (aq) (in acidic solution)
- Separate the equation into half-reactions:
MnO4 – → Mn2+
SO2 → HSO4–
- Balance the atoms and charges in each half-reaction:
MnO4 – + 8H+ + 5e– → Mn2+ + 4H2O
SO2 + 2H2O + 2e– → HSO4– + 2H+
- Multiply each half-reaction by a coefficient so that the number of electrons gained in reduction equals the number of electrons lost in oxidation:
5SO2 + 2MnO4 – + 16H+ → 2Mn2+ + 5HSO4– + 8H2O
- Add the half-reactions together and cancel out any common terms:
5SO2 + 2MnO4 – + 16H+ → 2Mn2+ + 5HSO4– + 8H2O
- Check that the charge and atom balance is correct.
(c) H2O2 (aq) Fe2+ (aq) → Fe3+ (aq) H2O (l) (in acidic solution)
- Separate the equation into half-reactions:
H2O2 → H2O
Fe2+ → Fe3+
- Balance the atoms and charges in each half-reaction:
H2O2 + 2H+ + 2e– →
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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)?
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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)? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about 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)? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 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)? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about 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)? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)?.
Solutions for 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)? in English & in Hindi are available as part of our courses for Class 10.
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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)?, a detailed solution for 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)? has been provided alongside types of 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)? theory, EduRev gives you an
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