Question

balance the equation by ion electron method in acidic medium.

Answer 1

Oxidation Half Reaction:

I - --------------> I 2  ____________ (1)

Reduction Half Reaction:

MnO4^1- -------------> Mn^2+_________(2)

Balancing oxidation half reaction:

2 I- - ----------> I2 +2e- ( As single iodide I- loses one electron to convert into I neutral. So 2 iodide will need to lose 2 electrons to convert into I2.)

Balancing reduction half reaction:

As Oxygen (O) is missing on Right hand side of equation 2, so to balance it we use 4 molecules of water as 4 oxygen can be obtained through this. So,

MnO4^1- +5e- -------> Mn^2+ + 4 H20 ( As in MnO4- oxidation number of Mn is +7 and on right hand side it is +2. it means it has gained five electrons.)

Now we have to balance the oxygen on left hand side of above equation because we wrote a source of oxygen on right side (H20) which is also a source of hydrogen. However there is no H on left side, so we simply add 8H+ on left side.

Mn04- + 5e- + 8H+ --------> Mn^2+ + 4H20________(3)

Now we multiply equation 1 by 5 and equation 3 by 2 to equalize the loss and gain of electrons. So the equations will be modified as:

      10 I^-1                          --------> 5 I2 + 10e-

2MnO4^-1 + 16H+ + 10e- ------> 2Mn^2+ + 8H20

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10  I^-1 + 2MnO4^-1 + 16H+---------> 2Mn2+ + 8H20

The loss and gain of electrons is 10. So they can be cancelled.  

The net equation (balanced) is thus obtained.