c)If 10.0 ml of 3.0 molar AgNO3 solution is added to the half cell on the right what will happen to the cell voltage? Explain.
d)If 20.0 ml of distilled water is added to both half cells, the cell voltage decreases. Explain

_____________________________________
| |
| ______Volmeter______ |
| | | |
| | _K+____NO3_ | |
| | | Salt bridge | | |
| | []Cd | | | | []Ag | |
| | _[] __ |_ | | |_ []___ | |
| | [] | | | | [] | |
| | | | | |
| | _______ | |_______|
100 ml 10ml
1 M Cd(NO3)2 1M AgNO3

My net equation is
Sr(s) --> Sr(2+) + 2e- 2.89V
Mg(2+) + 2e- --> Mg(s) -2.37V
E cell = 0.52V
Net : 2Ag+ Cd(s) --> 2Ag(s) + Cd(2+)

Sr I try to draw the picture by line and I forgot to review before post so it look like this.

It would be a good idea to draw a picture and attach it (click the "Go Advanced" button if it's visible. If not, there's something in "My Profile" that you can change to get it.) In any case, I clicked "Quote User" and I can see the diagram a bit more easily.

The half-cell on the right is Ag \rightarrow Ag^+ + e^-. If you add Ag^+ to it AgNO_3, it will inhibit the reaction so it will lower the cell voltage.

If you dilute the solutions, there are fewer ions to react per unit volume (the concentration drops). The EMF is a function of concentration, so the voltage will drop.

Wait... why did you use the half equation of Strontium and Magnesium up there?