So you need to draw a voltaic cell... you can draw this by hand, or you can draw it electronically. DO NOT just find one online and copy-paste it into your hand-in, if that's all I wanted I would have pasted this {voltaic cell} into the hand-in myself before I posted it.
The part that causes some confusion is the "...write a correctly balanced net ionic equation for the spontaneous process..." Don't over-read that! The metal cation solutions you were using in lab were probably nitrate salts, but nitrate (or whatever anion might have been present) was a spectator in all of your reactions. Net ionic equations are actually easier than full-formula equations because they're not cluttered up with with a bunch of extra stuff, and net ionic equations actually just describe the CHEMISTRY that's happening rather than distracting you with a bunch of spectator ions and species.
Now, I'm not going to write out a net ionic equation that's the exact answer to one that you have to write, but here's an example. Let's say I made up the voltaic cell Fe|Fe+3||Cd+2|Cd with the black/negative lead of my meter hooked up to the Fe(s) electrode and the red/positive lead connected to the Cd(s). The potential I measure is -0.32V. Because the measured potential is negative, the cell is running backwards, so the spontaneous cell reaction is Cd|Cd+2||Fe+3|Fe. Translating that into a reaction, we can write the two half-reactions as:
Cd(s) <=> Cd+2(aq) + 2 e-
3e- + Fe+3(aq) <=> Fe(s)
Adding those up gives the overall (or "net") reaction:
3Cd(s) + 2 Fe+3(aq) <=> 2 Fe(s) + 3 Cd+2(aq)
{Remember to multiply each half-reaction by an appropriate integer to make all the electrons cancel...}
That's a "correctly balanced net ionic equation for the spontaneous process" in this case. Now go do that for all the cells you measured in the experiment.
