Solveeit Logo
Login

Question

Question: Balance the following redox reactions by the ion-electron method in the basic medium. \({ MnO }_{ ...

Balance the following redox reactions by the ion-electron method in the basic medium.
MnO4−(aq)+I−(aq)→MnO2(s)+I2(s){ MnO }_{ 4 }^{ - }{ (aq)+{ I }^{ - }{ (aq)\rightarrow { MnO }_{ 2 }{ (s)+{ I }_{ 2 }{ (s) } } } }

Explanation

Solution

Hint : Ion-electron method: It is also known as the half-reaction method, the redox reaction is isolated into two-half conditions: one for reduction and one for oxidation. Each one of these half-reactions is balanced independently and then combined to give a balanced redox reaction.

Complete step by step solution :
There are following steps to balance a redox reaction by the ion-electron method in a basic medium which is explained below:
Step 1: The two half-reactions intricate in the given reaction are:
Oxidation: I2(aq)→I2(s){ I }_{ 2 }{ (aq)\rightarrow I }_{ 2 }{ (s) }
Reduction: MnO42−(aq)→MnO2(aq){ MnO }_{ 4 }^{ 2- }{ (aq)\rightarrow MnO }_{ 2 }{ (aq) }

Step 2: Balance I in the oxidation half-reaction, we have
2I−(aq)→I2(s){ 2I }^{ - }{ (aq)\rightarrow I }_{ 2 }{ (s) }
Now, to balance the charge we have to add two electrons in RHS of the reaction.
2I−(aq)→I2(s)+2e−{ 2I }^{ - }{ (aq)\rightarrow I }_{ 2 }{ (s)+2e^{ - } }

Step 3:The oxidation state of Mn has reduced from +7 to +4 in the reduction half-reaction.
Thus, three electrons are added to the LHS of the reaction.
MnO4−(aq)+3e−→MnO2(s){ MnO }_{ 4 }^{ - }{ (aq)+{ 3e }^{ - } }{ \rightarrow MnO }_{ 2 }{ (s) }
Now, to balance the charge we have to add four ions to the RHS of the reaction as the reaction is taking place in a basic medium.
MnO4−(aq)+3e−→MnO2(s)+4OH−{ MnO }_{ 4 }^{ - }{ (aq)+{ 3e }^{ - } }{ \rightarrow MnO }_{ 2 }{ (s)+4OH }^{ - }

Step 4: In this reaction, there are six O atoms on the RHS and four O atoms on the LHS. Therefore, two water molecules are added to the LHS.
MnO4−(aq)+2H2O+3e−→MnO2(s)+4OH−{ MnO }_{ 4 }^{ - }{ (aq)+{ 2H }_{ 2 }{ O+3e }^{ - } }{ \rightarrow MnO }_{ 2 }{ (s)+4OH }^{ - }

Step 5: Now, equalize the number of electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, we have:
6I−(aq)→3I2(s)+2e−{ 6I }^{ - }{ (aq)\rightarrow 3I }_{ 2 }{ (s)+2e^{ - } }
2MnO4−(aq)+4H2O+6e−→2MnO2(s)+8OH−{ 2MnO }_{ 4 }^{ - }{ (aq)+{ 4H }_{ 2 }{ O+6e }^{ - } }{ \rightarrow 2MnO }_{ 2 }{ (s)+8OH }^{ - }

Step 6: Now, by adding the two half-reactions, we have the net balanced redox reaction as:
6I−+2MnO4−(aq)+4H2O→2MnO2(s)+3I2(s)+8OH−{ 6I }^{ - }{ +2MnO }_{ 4 }^{ - }{ (aq)+{ 4H }_{ 2 } }{ O\rightarrow 2MnO }_{ 2 }{ (s) }{ +3I }_{ 2 }{ (s)+8OH^{ - } }

Note : The possibility to make a mistake is to balance a charge in oxidation half-reactions by adding electrons in the product, not a reactant, and in reduction half-reaction, the charge is balanced by adding electrons as a reactant.