How do you predict redox reactions?

Look at the two possible oxidation states of iron and tin.  Iron is either 2+ or 3+ while tin is either 2+ or 4+.  That should tell you that the redox will occur between the iron and the tin.

Fe3+  + e- ---> Fe2+
Sn2+ ---> Sn4+  +  2 e-

2 Fe3+ + 2 e- + Sn2+ ---> 2 Fe2+  + Sn4+   + 2 e-

2 Fe3+  + Sn2+ ---> 2 Fe2+  + Sn4+

if all of the products and reactants are soluble and all of the ions cancel out, the result is that it does not react.  you can only write a net ionic equation when there is some change taking place on either side of the equation because that's what it is for, to show the change that took place in the reaction

OK...in order to answer your question fully we need to consider two points.

a) What would the products be if the reaction had to occur?

Fe3+ is reduced to Fe2+ according to the half-equation:
Fe3+ + e Fe2+

Sn2+ is oxidised to Sn4+ according to the half-equation:
Sn2+  Sn4+ + 2e

However, since the number of electrons involved in both cases has to be the same, then we have to multiply the first equation by 2, to obtain:
2Fe3+ + 2e  2Fe2+

Summing up the two half-equations:
2Fe3+ + Sn2+  2Fe2+ + Sn4+

b) Should the reaction occur?
Just because you have determined the redox products does not necessarily guarantee that the reaction will occur. To determine this you need electrode potentials. In this particular case they are:

Fe3+ + e > Fe2+   E = +0.77V
Sn4+ + 2e > Sn2+  E = +0.154V

Now, a positive electrode potential implies that the equilibrium prefers to go to the right, so Fe2+ and Sn2+ are the preferred states for both half-equations. However, Fe3+ prefers to be reduced (0.77) much more than Sn2+ does not prefer to be oxidised (0.154), so the Fe3+ will win and the reaction should occur.

Please notice that I say SHOULD occur because this is merely a thermodynamic prediction. It is no guarantee that the reaction will, from a kinetic perspective, be fast enough or even proceed at all.