Question

Balance the following redox reaction by oxidation number method. 3

MnO2 + Br– → Mn2+ + Br2 + H2O (Acid medium)​

Answer 1

Answer:

Explanation:

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)