BALANCE THE REDOX REACTION USING HALF CELL METHOD IN AN ACIDIC MEDIUM
CHCl3 + MnO4- = Cl2 + CO2 + Mn+2
MnO4^- + 8H^+ +5e = Mn^2+ + 4H2O
2CHCl3 + 4H2O==> 3Cl2 + 2CO2 + 10 H^+ + 10e
Multiply equation 1 by 10; equation 2 by 1 and add,
To balance the redox reaction using the half-cell method in an acidic medium, follow these steps:
1. Write the unbalanced equation: CHCl3 + MnO4- = Cl2 + CO2 + Mn+2
2. Separate the half-reactions:
a) Oxidation half-reaction (loss of electrons): CHCl3 → Cl2
b) Reduction half-reaction (gain of electrons): MnO4- → Mn+2
3. Balance the atoms other than hydrogen and oxygen in each half-reaction. Start with the most complex species and balance one element at a time.
a) Oxidation half-reaction (Cl2 formation):
1) The number of chlorine (Cl) atoms on the reactant side is 3, and on the product side is 2. Add a coefficient of 3 in front of Cl2 in the product side to balance chlorine atoms. The equation becomes: CHCl3 → 3Cl2
b) Reduction half-reaction (Mn+2 formation):
1) Start by balancing the manganese (Mn) atoms. The number of Mn atoms on the reactant side is 1, and on the product side is 1. No coefficient is needed for Mn.
2) Balance the oxygen (O) atoms by adding water (H2O) molecules. On the reactant side, there are 4 oxygens, so add 4 H2O molecules on the product side. The equation becomes: MnO4- → Mn+2 + 4H2O
4. Balance the hydrogen (H) atoms by adding H+ ions to each side of the equation. The number of hydrogen atoms on the reactant side is 3 from CHCl3 and 8 from the 4 H2O molecules, making it 11 hydrogen atoms. So, add 11 H+ ions to the product side of the oxidation half-reaction:
CHCl3 + 11H+ → 3Cl2
5. Balance the charge in each half-reaction by adding electrons (e-):
a) Oxidation half-reaction: In the oxidation half-reaction, the overall charge on both sides is balanced, so no electrons are needed.
b) Reduction half-reaction: The charge on the reactant side is +7, and on the product side is +2. So, add 5 electrons to the reactant side to balance the charge. The equation becomes: MnO4- + 5e- → Mn+2 + 4H2O
6. Multiply each half-reaction by a suitable factor, if necessary, to make the number of electrons in the oxidation and reduction halves equal.
a) Multiply the oxidation half-reaction by 5:
5CHCl3 + 55H+ → 15Cl2
b) Multiply the reduction half-reaction by 3:
3MnO4- + 15e- → 3Mn+2 + 12H2O
7. Add the balanced half-reactions.
5CHCl3 + 55H+ + 3MnO4- + 15e- → 15Cl2 + 3Mn+2 + 12H2O
8. Cancel out the electrons on both sides of the equation.
5CHCl3 + 55H+ + 3MnO4- → 15Cl2 + 3Mn+2 + 12H2O
Your balanced redox reaction using the half-cell method in an acidic medium is:
5CHCl3 + 55H+ + 3MnO4- → 15Cl2 + 3Mn+2 + 12H2O