In order to answer this question, it is important to realize that if the equation is properly balanced then the #e- lost is equal to the #e- gained. It looks to be balanced, so we can use oxidation numbers to determine what has been oxidized (lost e-) and reduced (gained e-) or really just one or the other.
Atoms in solid tin, Sn(s), have an oxidation state of zero. On the other side of the equation, notice that tin is now an ion with a +2 oxidation state. Each atom of tin has lost 2e- in order to go from a 0 to a +2. Since there is a coefficient of 3 (3Sn(s) AND 3Sn2+(aq)), that must mean 6 total e- were lost. So 6 e- must have been gained. In the nitrate ion, NO3-, the N has an oxidation state of +5 because each O is -2 and the overall charge of the nitrate ion is a -1 (N + 3(-2)=-1). On the products side, NO(g) is produced. The N has an oxidation state of +2 since O again is a -2 (N +1(-2)=0). Each N atom then went from a +5 to a +2. That is 3 e- each times 2 (the coefficient) gives us 6e- gained!!
So, we can see that the #e- lost = #e- gained.
I hope this helps. :)
