CHCl3 + MnO4- = Cl2 + CO2 + Mn+2

Balance the above with half reactions in acidic medium.

For the above reaction, can you please show the half reactions...which is getting oxidised...both C and Cl's oxidation # are changing showing both getting oxidised,.how is this possible? Can u please do the steps...I may have something wrong Many thanks

I gave you the two balanced half reactions yesterday. Why do I need to this again? C goes from +2 on the left to +4 on the right. Cl is -1 on the left and 0 on the right. Both are oxidized. Nothing wrong with that.

Sure! Let's analyze the given reaction step by step to determine the half reactions and identify which species is getting oxidized.

First, let's write the unbalanced equation:

CHCl3 + MnO4- → Cl2 + CO2 + Mn+2

To make the balancing process easier, let's break down the reaction into half reactions. We will separate the oxidation and reduction processes:

Oxidation half-reaction: CHCl3 → Cl2
Reduction half-reaction: MnO4- → Mn+2

Now, we need to balance each half-reaction individually.

Let's start with the oxidation half-reaction. In CHCl3, carbon's oxidation state is +4, and in Cl2, it is -1.

To balance the change in oxidation state, we need to add electrons to the left side (reactants):

CHCl3 + 3e- → Cl2

Next, let's work on the reduction half-reaction. In MnO4-, manganese's oxidation state is +7, and in Mn+2, it is +2.

To balance the change in oxidation state, we need to add water (H2O) molecules and hydrogen ions (H+) to the right side (products):

MnO4- + 8H+ → Mn+2 + 4H2O

Now, we need to balance the number of atoms on both sides of each half-reaction.

For the oxidation half-reaction, we have one carbon atom on the reactant side and two chlorine atoms on the product side. To balance this, we can add a coefficient of 2 in front of the reactants:

2CHCl3 + 6e- → Cl2

For the reduction half-reaction, we have one manganese atom on the reactant side and one manganese atom on the product side. The number of hydrogen atoms is already balanced. Therefore, no additional coefficients are needed.

Now, we need to balance the number of electrons transferred in both half-reactions.

In the oxidation half-reaction, 6 electrons are transferred. To balance this, we can multiply the reduction half-reaction by a factor of 6:

6MnO4- + 24H+ → 6Mn+2 + 12H2O

Finally, we can combine both half-reactions to form the balanced equation:

2CHCl3 + 6MnO4- + 24H+ → 6Mn+2 + 12H2O + Cl2

In this reaction, both carbon and chlorine are being oxidized. Carbon's oxidation state increases from +4 to +2, and chlorine's oxidation state increases from -1 to 0. So, both carbon and chlorine are undergoing oxidation in this reaction.

I hope this explanation helps! Let me know if you have any further questions.