Confused about electrochem potentials

[Mp07d]

I used to think I had electrochem on lock ( at least for solving for potentials ) but it wasnt when I saw this question on test 7

Cu2+ + 2e- --> Cu(s) Eo = +0.34 V
2H20 --> O2 + 4H+ + 4e- Eo = -1.23 V

What is the Eo(cell) for the reaction shown in the following equation?

2Cu2+ + 2H20 --> 2Cu(s) + O2 + 4H+

A) -0.89 V
B) +0.55 V
C) +1.57 V
D) + 1.91 V


That I got completely confused. I used to think you always did Cathode-Anode ( which in this case wouldnt it be .34-(-1.23)??) But apparently here it is Cathode + Anode. I'm..pretty confused now. Can someone please clear this up for me. Thanks

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[Rabolisk]

What exactly does Cathode - Anode mean?

 

[KindOfMaroon]

Don't you need to double that potential from the Cu rxn?

edit: also, here's how I do these:

I look at the two half-reactions for stoichiometry and direction (like, which side are the electrons on). Then, I look in the final reaction they want me to solve for. Are either of the half-reactions flipped? Is the stoichiometry different?

If the half-reaction gets flipped in the rxn you're trying to solve, then simply make the potential negative from the half-cell value.
If the stoichiometry is doubled, then double the potential.

Then you just sum all the potentials.

Does that make sense?

 

[Mp07d]

cathode minus anode

and no you dont double the values for E cell potential. All you ever do is flip the sign.

 

[Rabolisk]

But what do you mean by cathode minus anode exactly?

 

[mcgill2012]

yeah i actually saw this question today and it tripped me up too.
there are two equations you can use to find cell potential:

EMF = Ecathode - Eanode
i think this is the one you were talking about. you can only use this when both of your potentials are reduction potentials. if you look at the given half reactions, the first is an oxidation and the second is a reduction - so you would have to flip the second reaction and flip the sign of the second potential. if you do this, you see that it is a more positive reduction potential. since the overall reaction is the sum of the two half reactions in the directions given, we therefore have an electrolytic reaction, because the cathode has a smaller reduction potential than the anode. this is a positive delta G and negative EMF, which eliminates all the others. Ecathode - Eanode = 0.34- (+1.23) = -0.89

alternatively the other equation (which was used in the solution) is
EMF = Ered + Eox
in which Ered is the reduction potential of the species getting reduced, and Eox is the oxidation potential of the species getting oxidized. since your half reactions are given in the directions taken in the overall reaction, and one's a reduction and ones an oxidation, you dont have to flip any of the signs.
EMF = Ered + Eox
= 0.34 + (-1.23) = -0.89

i know i rambled a bit but i hope that helps!

 

[Mp07d]

so basically I should assume that the Evalue they give me is the reduction potential and if I have an oxidation going on I should flip the value to get the real oxidation value?

 

[mcgill2012]

dont assume the value they give you is a reduction potential.
look at the half reactions. if electrons are being consumed, the potential given is a reduction potential.
if electrons are being produced, the potential given is an oxidation potential.

 

[Rabolisk]

That's backwards. You should assume that the Evalue they give you is for the reaction shown, and you should flip the value to get the reduction value if they give you an oxidation half-reaction.