Balancing of Redox Reactions with solved example.

Redox reactions are reactions involving reduction and oxidation. Now what is oxidation and reduction.
Oxidation: is a process involving a loss of electrons(electrons are negative so loss of electrons will lead to increase in electropositivity).
Reduction : Is a process involving gain of electrons i.e increase in electrons will reduce electropositivity and increase electronegativity.

Balancing of Redox Reactions.

In alkaline solutions hydroxide(OH-) and water(H2O) are used to balance oxygen and hydrogen  respectively. For each oxygen molecule 2OH- ions are added to the side of the half equations that is deficient(lacking) oxygen and one mole of water(H20) is added to the opposite side to balance the hydrogen.
In acidic solution H+ and H20 are used to balance oxygen and hydrogen.
Always split your redox reactions to half equations, one half showing oxidation, the other showing reduction.

 Example.

Balance Cr2O7²- + NO2- -->Cr³+ + NO3-. This is an acidic reaction so we will use H+ and H2O to balance it.

First of all split the reaction into half equation.
Cr2O7²-  -->Cr³+ (reduction, Chromium reduces from +6 to +3,if you find the oxidation numbers, gain of electrons reduces the positivity of an element).
NO2-  -->NO3- (oxidation, nitrogen increases from +3 to +5,means it loses electrons, read my post on finding oxidation number).
Step2: Balance any atom other than oxygen and hydrogen first. Work with one half reaction at a time.
  Balance Cr.
Cr2O7²- -->Cr³+,

To balance Cr at the right hand side(RHS), we need to put 2 at the RHS, to balance it chemically, so we have
Cr2O7²- -->2Cr³+
Chromium is balanced. Now lets face oxygen.
We have 7moles of oxygen at the LHS and none at the RHS, to balance a mole of oxygen is with 1H2O,so 7moles of oxygen will need 7H2O.
Cr2O7²- -->2Cr³+ + 7H2O
Now oxygen is balanced but hydrogen is 14 at RHS and zero at LHS. so we balance a mole of Hydrogen with H+, so 14mole will be 14H+
Cr2O7²- +14H+ -->7H20 + 2Cr3+.
Mass balance achieved for the first half reaction.All that is remain is charge balance.
  To balance the charge, find the net charge on each side of the half reaction and add as many electrons(e) as needed to the positive side, so as to equal the negative side.
Net charge : LHS half reaction is +12 ( 14H+ and -2 from Cr2O7²-)
Net charge : RHS is +6( +6 due to two Chromium iii ions (2Cr³+ = 2×+3=6)
By adding 6electrons (6e-) to the more positive side(LHS) ,the net charge will be balanced.
6e- + 14H+ + Cr2O7²- -->2Cr³+  +7H2O.

Second half reaction.

NO2- -->NO3-. nitrogen is already balance, so lets go to oxygen.
Balance Oxygen : H2O+ NO2- -->NO3-.
Balance Hydrogen : H2O + NO2- -->NO3- +2H+.
Mass balance achieved.
Net charge :LHS =-1
Net charge : RHS = +1 (-1 from NO3- +2 from 2H+=-1+2=+1).
So by adding 2e- to the RHS will balance the equation.
H2O+NO2- -->NO3- +2H+ +2e-
Examination of the two half reactions shows:
6e-+14H+ +Cr2O7²- -->2Cr³+ +7H2O.
H2O + NO2- --> NO3- +2e- +2H+.

You will see 6e- in first half equation and 2e- in the second half equation . To balance it you multiply first equation by 2 and the second by 6 or you can use 1 and 3(that's 2 and 6 in lowest term).
Hence we have:
[6e- + Cr2O7²- -->2Cr³+ +7H20]×1
[ H2O+NO2- -->NO3- +2e- +2H+] ×3
3H2O + 3NO2- -->3NO3- +6e- + 6H+.
Now that the electrons are equal in both equations you cancel them out.
Now we subtract any double H+ and H2O,
So we have 14H+ - (6H+)=8H+
7H2O-3H2O=4H2O.
So you can now join the equations.
Cr2O7²- +3NO2- +8H+ -->2Cr³+ +3NO3- +4H20.
Net charge at both sides =+3.

To balance this in alkaline or basic solution add equal number of OH- to the H+. This will form water. Then you subtract the water from both sides and the remaining water takes at that side.
So we will add 8OH- to the 8H+= 8H20,so if you subtract 8H2O from 4H2O. we have 4H2O at the LHS.
So in basic solution it will be
4H2O + Cr2O7²-  +3NO2- -->2Cr³+ +3NO3-.
YOU CAN ALSO READ MY POST ON OXIDATION NUMBER here