What's the difference between e.m.f and terminal p.d ?

The emf of a cell is the emf measured with no current flow,The terminal potential difference is equal to the emf only if the load resistance is infinite.Consider it this way.Imagine a series circuit made up with a fixed emf ,say 2v,(due to the cell chemistry),then a small resistor(R1) ,say 0.5 ohms(to act as the internal cell resistance)and a further resistance which can take values(R2) at your choice of ,say 1,10,100 and 100 ohms.Now I do a calculation and then follow on with the others, so you get the picture.I take the case R1=0.5 ohm. R2=10ohm.current flow is v/(R1 +R2)=2/10.5 =0.19048a.Now calculate voltage drop over R1 and R2 separately as V = I x R .I get voltage drops of 0.09523 over R1 and 1.9048 over R2.adding these two potential drops we get 2.00003 volts ,which add up to the emf of the cell(it would add up exactly to 2 had I used full accuracy.I suggest it would help you get a handle on this if you do the calculations for the other values of R2.Good luck,keep at it!