<p>Hi,
this was a question on one of the Barron's practice problems but I has no idea how to solve it.
Essentially the question provided this picture: [url=<a href="http://www.flickr.com/photos/63434575@N03/5773651546/%5Dvoltaic">http://www.flickr.com/photos/63434575@N03/5773651546/]voltaic</a> cell | Flickr - Photo Sharing<img src="sorry,%20my%20Paint%20skillz%20suck" alt="/url">.
and asked to find the anode half reaction.
Can someone please explain to me how I'm supposed to determine the oxidation reaction from the picture (the book explanation was useless).</p>
<p>If I can not figure something out and it has free response, I put, " The secret to life is love and 42." :P<br>
/random thing I do.</p>
<p>This is from the redox chapter in barron’s right? (i just did it today) I think they left out an arrow or something in the diagram, since there should be one indicating the flow of electrons from the zinc electrode to the copper. Otherwise you would need an activity series.</p>
<p>Yes it is!
So using the activity series, would you determine whichever one is more easily oxidized and reduced to determine what is at the cathode/anode?</p>
<p>Well both the anode (zinc) and cathode (copper) have an ionic state of 2+, so they both give/lose 2 electrons. Since zinc is the anode, remember the mnemonic an ox, red cat, so oxidation occurs at the anode. Also, remember Leo says Ger, for lose electrons oxidation, gain electrons reductions. So, the zinc will be losing electrons, and since you know it must lose 2 electrons, the electrons will be on the right side of the reaction. Therefore, the final reaction will be Zn –> Zn2+ + 2e-. </p>
<p>I hope this helps!!</p>