Selecting Anode & Cathode for Cu2(SO4) & Fe(SO4) Flow Battery

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IntegrateMe
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A graduate student is trying to develop a flow battery based on aqueous solutions of 1 M Cu2(SO4) and Fe(SO4) (recall that the sulfate is divalent, e.g. (SO4)-2 and assume complete solubility).

Given a chart like this : CHART

How would I know that I need to select Copper (I) and Iron (II) as my anode and cathode, respectively?

I haven't taken chemistry before, so if anyone can explain in a concise way I'd really appreciate it.

Thanks!
 
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You can't select Cu(I) nor Fe(II) as anode or cathode - anode and cathode have to be a conductive solid (or liquid), you can't make them out of a single ion.
 


Sorry, I'm not really sure how to respond. The answer for this question on my last exam was:

Anode: 2Cu+ + 2e- → 2Cu
Cathode: Fe → Fe2+ + 2e-
Overall: Cu2(SO4) → Fe(SO4) + 2Cu

I am just confused as to why the anode and cathode were selected to be those two, as opposed to some other form of copper or iron...
 


These are not cathode and anode material, but cathode and anode reactions. If you don't know the difference you have a lot to learn, that's not a thing I can help you in a single post.

Look for any general chemistry material describing the cells, redox reactions and reactivity series.
 


This is supposed to be a pretty easy question, are there any tricks you can offer that may help me solve other ones (i.e. the cathode is usually the reaction we flip and make the potential negative)? Also, I'm still confused as to why we chose Copper(I) and Iron(II) instead of some other form...
 


Cu(I) and Fe(II) are part of the two solutions that you are given, that is why you are using them. As for your anode and cathode, I believe you must choose so that your overall E° is negative, as this is an electrolytic cell and the E° is always negative in an electrolytic cell. At least that's what I remember. Anyone care to clarify? My chemistry is a little rusty.