Question

how to balance equation in acidic medium

Answer 1

with the help of ico-eletronic method..

,,,,,,,,,,,,,,,,,,,,,,,, process of method,,,,,,,,,,,,,

First we must identify what's being oxidized and what's being reduced. Start by assigning oxidation numbers to everything and see what elements' oxidation numbers increase (and are being oxidized) and which ones decrease (and are being reduced). 

The oxidation number of Mn in the permanganate ion is +7. When Mn appears on the other side of the equation it has a charge of +2, which means its oxidation number is ALSO +2. Manganese is being reduced. 

The oxidation number of carbon in oxalic acid is +3. The oxidation number of carbon in CO2 is +4, so carbon is being oxidized. 

Now write half-reactions showing oxidation and reduction separately. Add the appropriate number of electrons to each half-reaction. For the oxidation half-reaction, each carbon atom must lose one electron; however, there are two carbon atoms in one molecule of oxalic acid. Therefore, add 2 electrons to the right side of the oxidation half-reaction. For the reduction half-reaction, each manganese must gain 5 electrons in order to be reduced from +7 to +2, so add 5 electrons to the left-hand side of the reduction half-reaction. 

Oxidation: H2C2O4 --> 2 CO2 + 2 e- 

Reduction: MnO4- + 5 e- --> Mn2+ 

Now it's time to balance the hydrogens and oxygens that come along for the ride. Since the reaction is taking place in an acidic medium, balance the oxygen atoms by adding water molecules to the opposite side of each half-reaction, then compensate for the hydrogen atoms by adding H+ ions as needed. When you are done, each half-reaction should show the same number and type of atoms on each side. 

Oxidation: H2C2O4 --> 2 CO2 + 2 e- + 2 H+ 

Reduction: 8 H+ + MnO4- + 5 e- --> Mn2+ + 4 H2O 

Now you've probably noticed that the number of electrons being released in the oxidation half-reaction is not equal to the number of electrons being absorbed during the reduction half-reaction. We need to rectify this. 2 won't go into 5, and the lowest whole number that both will go into is 10. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 so that the number of electrons is the same: 

Oxidation: 5 H2C2O4 --> 10 CO2 + 10 e- + 10 H+ 

Reduction: 16 H+ + 2 MnO4- + 10 e- --> 2 Mn2+ + 8 H2O 

Now recompile the half-reactions into one equation: 

16 H+ + 2 MnO4- + 10 e- + 5 H2C2O4 --> 2 Mn2+ + 8 H2O + 10 CO2 + 10 e- + 10 H+ 

Eliminate or reduce anything that appears on both sides. This leaves: 

6 H+ + 2 MnO4- + 5 H2C2O4 --> 2 Mn2+ + 8 H2O + 10 CO2 

I hope that helps. Good luck!