# Chemistry Class 11 NCERT Solutions Chapter 8 Redox Reactions Part 8

Q: 18. Balance the following redox reactions by ion-electron method

(A) (In basic medium)

(B) (In acidic solution)

(C) (In acidic solution)

(D) (In acidic solution)

(A) Step 1: The two half reactions involved in the given reaction are:

Oxidation half reaction:

Reduction half reaction:

Step-2: Balancing I in the oxidation half reaction, we have:

Now, to balance the charge, we add to the RHS of the reaction.

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

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

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

Step 5: Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:

Step 6: Adding the two half reactions, we have the net balanced redox reaction as:

(B) Following the steps as in part (A) we have the oxidation half reaction as:

And the reduction half reaction as:

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:

(C) Following the steps as in part (A), we have the oxidation half reaction as:

And the reduction half reaction as:

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

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

And the reduction half reaction as:

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

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