Suppose you have the following voltaic cell: $\ce{Sn_{(s)}|Sn^{2+}_{(aq, 1.0 M)}||Cu^{2+}_{(aq, 1.0 M)}|Cu_{(s)}}$ and the salt bridge is $\ce{KNO_3}$. What I don't understand is why you need to have $\ce{Sn^{2+}}$ ions initially in the $\ce{Sn}$ half cell. As $\ce{Sn}$ is oxidized, electrons are supposed to travel through the wire connecting the two half cells and reduce the $\ce{Cu^{2+}}$ ions in solution, but if we have $\ce{Sn^{2+}}$ ions initially in the $\ce{Sn}$ half cell, the electrons might just as well reduce the $\ce{Sn^{2+}}$ ions. Why would the electrons travel all the way through the wire when there is $\ce{Sn^{2+}}$ ions right next to them that could be reduced?
I also don't understand why the $\ce{Cu}$ cathode has to exist. All we need is $\ce{Cu^{2+}}$ ions in solution that will be reduced, so why can't we just dip the wire directly into $\ce{Cu^{2+}}$ solutions. As the reaction proceeds, a $\ce{Cu}$ cathode will be created anyways as $\ce{Cu}$ metal precipitates on the wire.
I am obviously misunderstanding something, because according to my understanding the following voltaic cell should work: In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire. The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution. The $\ce{Sn}$ gets oxidized and produces $\ce{Sn^{2+}}$ ions in the water. The electrons travel through the wire and reduce the $\ce{Cu^{2+}}$ ions in the cathode solution. The anode solution will become more positive and the cathode solution will become more negative, so we still need a salt bridge.