2
$\begingroup$

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.

Improve this question
$\endgroup$
3
3
$\begingroup$

A voltaic circuit needs to be a complete circuit first an foremost. So, to answer your 2 main concerns, think of a voltaic cell as a electric circuit:

In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire.

Water is not a good conductor of electricity, dissolved ions are needed, such as the $\ce{Sn^{2+}}$ ions are needed to facilitate the movement of the electric current in the first place. Also, the presence of the ions allows the overall redox reaction be spontaneous to provide the electric current (ChemWiki).

The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution

According to the ChemWiki Voltaic Cells, the metals plates provide a surface for the reactions to occur on, a wire just submersed into the solution provides a very small surface area for the reaction to occur.

A diagram from the UC Davis ChemWiki page Electrochemical Cells helps explain this (the example uses zinc instead of tin, but the principle is the same):

enter image description here

Improve this answer
$\endgroup$
3
  • $\begingroup$ Ahhh you are totally right about the second point, the cathode provides a larger surface area. As for the first point, can you please elaborate why you need $Sn^{+2}$ ions dissolved for the reaction to take place? You say they are needed to keep the current going, but I don't see why this is. $\endgroup$ – Joshua Benabou May 3 '15 at 22:24
  • $\begingroup$ Water alone is a poor conductor of electricity (hence would 'break the circuit'), the charged ions provide a means for the circuit to be complete - the questions and answers of Why can't pure water conduct electricity since it can be reduced at cathode and oxidised at anode? provide some insight - in short, to 'get around' water's high electrical resistance, charged ions are needed to complete the circuit. $\endgroup$ – user15489 May 3 '15 at 22:40
  • $\begingroup$ But you get ions from the salt bridge which keep the electrode solutions electrically neutral, so why do you need the $Sn^{+2}$ ions? And is it not true that the $Sn^{+2}$ ions might accept the electrons from the $Sn$ electrode? $\endgroup$ – Joshua Benabou May 4 '15 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.