[beheada]

First off, I was wondering if there was anything in particular I could read to get a jump start on organic, since I'm taking it next year. I just got done with my first two chem classes and labs and wound up with 4 As. But, on that note, I just took my final and the second to last question about killed me. I sat there for at least half an hour trying to figure this out:

Calculate the potential and write balanced equations for the spontaneous redox reaction of gold and copper. Given was:

Au⁺³(aq) + 3e^- ---> Au(s)     E=1.50 v
Cu⁺²(aq) + 2e^- ---> Cu(s)      E=0.34 v

Then there were several other half-reactions showing electron transfer, such as MnO4, Br2, NO + H but I can't remember them all to post. Any help would be appreciated? I'm pretty sure this is the only question I missed, that's why it's vital I must find ze answer. If that isn't enough information, I can see about emailing my professor for the exact question. 

[Albert]

Well, without the other half-reactions it seems a too broad question. 

A possible reaction, according to what you wrote, is

Au3+ + 3 NO + 3 H2O -> Au + 3 HNO2 + 3 H+

However, I just compared the two standard potentials and balanced the whole reaction. 

[Borek]

write balanced equations for the spontaneous redox reaction of gold and copper.

There is only one redox reaction between these that can be spontaneous. Think about in simple terms of reactivity series.

Au⁺³(aq) + 3e^- ---> Au(s)     E=1.50 v
Cu⁺²(aq) + 2e^- ---> Cu(s)      E=0.34 v

What about Nernst equation?

[Alberto_Kravina]

Calculate the potential and write balanced equations for the spontaneous redox reaction of gold and copper. Given was:

Au+3(aq) + 3e- ---> Au(s)     E=1.50 v
Cu+2(aq) + 2e- ---> Cu(s)      E=0.34 v

Just calculate DeltaE⁰ (assuming that the reaction happens at standard conditions) ?DeltaE⁰=E⁰(Reduction) - E⁰(Oxidation)
If the reaction doesn't take place at Standard conditions you have to use the Nernst Equation to calculate the potential of each half-cell

Edit: Borek was about 40 seconds faster

[Borek]

Seems the question is ambiguous - Albert got it different then I did.

[Borek]

Edit: Borek was about 40 seconds faster

Albert was even faster

[Albert]

Calculate the potential and write balanced equations for the spontaneous redox reaction of gold and copper.

Not being English my mother language, I might be the last person to say this, but, still, I think there's a mistake in this sentence.
There're two possibilities: the former is that it should be 'reactions', instead of its singular; the latter is 'with', instead of 'and'.

[Alberto_Kravina]

Seems the question is ambiguous - Albert got it different then I did.

Yeah - However I think that beheada meant the reaction between Cu⁰ and Au³⁺.
In this case it is pretty easy to predict the reaction, even without calculating deltaE⁰ since Gold cations aren't very  stable...

[beheada]

For clarity sake, I will email my professor and see if she'll email the question to me verbatum.

The way I read the question when I was in class was that there was a redox reaction that included gold and copper ions, one of which was reduced, the other oxidized. We haven't covered the Nernst equation, so I'm pretty sure that it doesn't involve that (this is pre-organic).

I was trying something like

Au(No₃)₃(aq)+CuBr₂(aq) + XH₂O ---> etc

I wrote a full page of different variants between Au, Br, Cu, MnO4, NO, and H but couldn't work out something that showed a redox of Au and Cu. Since I couldn't figure out the equation, I could figure out the E⁰. Tragedy to my almost perfect final exam. But, like I said, I'll email my prof when she releases the grades on Monday and ask her for the question and data sheet.

[Borek]

Ions will not react. It will be reaction between ion and metal.

[Alberto_Kravina]

There are two half-reactions.

1.)Au³⁺ + 3e^- ? Au⁰
2.)Cu⁰ ? Cu²⁺+ 2e^-

Try to balance these two equations. Just make sure that there is the same number of electrons on each side of the rection (so you just have to multiply the first half-reaction with 2 and the second with 3 and put these two half-reactions together)

Borek faster again....

[beheada]

There are two half-reactions.

1.)Au³⁺ + 3e^- ? Au⁰
2.)Cu⁰ ? Cu²⁺+ 2e^-

Try to balance these two equations. Just make sure that there is the same number of electrons on each side of the rection (so you just have to multiply the first half-reaction with 2 and the second with 3 and put these two half-reactions together)

Borek faster again....

Alright... somewhere on the page I scribbled all over I had:

if:
1) Au³⁺ + 3e^- ? Au⁰       E⁰= 1.50 v
2) Cu²⁺+ 2e^- ? Cu⁰         E⁰= 0.34 v

Reverse the second equation which reverses the sign, then multiply the Au by 2 and the Cu by 3, then combined the two equations: 1.50v + (-0.34v) to get 1.16v.  I left that there, but the equation just didn't look right so I didn't SQUARE the answer (which is how I normally proclaim that that is my definite final answer).

Since that didn't look right, I went off the deep end trying to incorporate all that other garbage.
I guess the equation I had wound up with was:

2Au(aq) + 3Cu(s) --> 3Cu(aq) + 2Au(s)      E⁰= 1.16v

[Alberto_Kravina]

I guess the equation I had wound up with was:

2Au(aq) + 3Cu(s) --> 3Cu(aq) + 2Au(s)

Check the charges! On one side Gold is ionic while it is atomic on the other side. Copper is atomic on the left side and ionic on the right side. What you wrote doesn't make sense...but if you add the correct charges to the Gold and copper ions the reaction turns correct!

[beheada]

I guess the equation I had wound up with was:

2Au(aq) + 3Cu(s) --> 3Cu(aq) + 2Au(s)

Check the charges! On one side Gold is ionic while it is atomic on the other side. Copper is atomic on the left side and ionic on the right side. What you wrote doesn't make sense...but if you add the correct charges to the Gold and copper ions the reaction turns correct!

Therefore:

2Au³⁺(aq) + 3Cu(s) --> 3Cu²⁺(aq) + 2Au(s)      E⁰ = 1.16v

Would be the correct equation? Maybe I'll get partial credit. Thanks for the help.

[Alberto_Kravina]

2Au3+(aq) + 3Cu(s) --> 3Cu2+(aq) + 2Au(s)     

Correct!

E0 = 1.16v

At standard conditions, yeah.