Redox reaction is also called Oxidation-Reduction Reaction. Sometimes the equations are easy enough to balance using the most common method we use for balancing the other reactions, but sometimes redox reactions can be complicated and difficult to figure out in 1 step. So we break down the equation and use half reactions to balance the equation.

Here is an example:

Balance the Redox Reaction in acidic solution:      BrO₃^-(aq)+SN²⁺(aq)--> Br^-(aq)+Sn⁴⁺(aq)

First of all, we can split the equation into two half reactions first and get:

BrO3-(aq)--> Br-(aq)

Sn²⁺(aq)--> Sn⁴⁺(aq)

Then we can use water H2O to balance the oxygen atoms O:

BrO₃^-(aq)--> Br^-(aq) + 3H₂O(l)

Now we use H⁺ to balance the above equation, since this is acidic solution (given in question):

BrO₃^-(aq) + 6H⁺(aq) --> Br^-(aq) + 3H₂O(l)

Then we add electrons to the left side in order to balance the charge and get this:

BrO₃^-(aq) + 6H⁺(aq) + 6e^- --> Br^-(aq) + 3H₂O(l)   (Half Equation 1)

Now it’s time to fix Sn²⁺(aq)--> Sn⁴⁺(aq) as well, using the same method:

Sn²⁺(aq)--> Sn₄⁺(aq) + 2e^-       

Here in order to cancel the electrons on both hal equations when adding them up, we need to multiply the above equation by 3:

3Sn²⁺(aq)--> 3Sn₄⁺(aq) + 6e^-     (Half Equation 2)  

We are almost done, just add Half Equation 1&2 together:

BrO₃^-(aq) + 6H⁺(aq) + 3Sn²⁺(aq) --> Br^-(aq) + 3H₂O(l)+ 3Sn₄⁺(aq)    

You see you fast that was? By the way, we use H⁺ to balance the equation because this reaction takes place in acidic solution. If there’s a question asking to balance a redox reaction in basic solution, then use H⁺ first, then use OH^- to cancel out the H⁺ later. Some people may ask why use H⁺ first all the time. But most people’s experience show that using H⁺ then OH^- is faster than just using OH^- (it takes more time to think if just use OH^- and it will easily mess up the answer.)