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Use the following table of standard electrode potentials:

Cu²⁺(aq) + 2e^- Cu     E°=+0.339V
Cu(OH)₂ + 2e^- Cu + 2 OH^-(aq)     E°=-0.224V
Cu(NH₃)₄²⁺ + 2e^- Cu + 4 NH₃(aq)     E°=-0.0510V

The Nernst equation is: E=E° - 0.0592/n*log Q

The equilibrium constant, K, for the following reaction at 298 K is: (K_b (NH₃)= 1.8*10^-5)

Cu(OH)₂(s) + 4 NH₃  Cu(NH₃)₄²⁺(aq) + 2 OH^-(aq)

A. 4.94 *10^-10
B. 1.40 *10^-6
C. 7.15 *10⁵
D. 2.02 *10⁹

I know that I have to eventually use nFE=RTln(K), and the only thing I've been having trouble with is getting the E. I assumed E°=-0.224+0.0510=-.173V. If I did this wrong, please let me know. Then, I assumed that the acid-base reaction of NH₃ would have some effect on the concentrations, so I found out that with an original solution of 1.0M NH₃, the NH₄⁺ and OH^- concentrations are both 0.00424M. After this, though, I was quite sure which values to use for finding the Q, which I could then use to find E and ultimately K.

If someone could direct me in the right direction, I'd greatly appreciate it! Thanks so much!

[Borek]

You can write separate Q for each four reactions.

You can calculate (separately) K constants for first three reactions.

Try to express Q of hydroxide dissolution in ammonia using first three Q, then plug in calculated K values.

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You can write separate Q for each four reactions.

You can calculate (separately) K constants for first three reactions.

Try to express Q of hydroxide dissolution in ammonia using first three Q, then plug in calculated K values.

I just tried doing that and got B as my answer, so I checked with the answer key, which says C?  If you want, I can show my work, although I was pretty sure that I did it correctly...

Just as a comment, this is the same answer you get as when you use nFE=RTln(K) with E= -.173V, which would then mean that we are assuming that all the concentrations are at 1.0 M. Do you think that I have to include different concentrations somehow?

[Borek]

Seems to me like the answer key is right - at least what I got was identical to C (assuming 0.0591 constant in the simplified Nernst equation).

It is about standard state, so you can assume all concentrations being equal 1M. You don't have to worry about ammonia reacting with water.

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Hmm. I had (assuming from top to bottom is Q₁ to Q₄, so Q₄ is the one I am trying to find) Q₄=Q₂/Q₃. I assume I'm calculating the K's wrong then.

Here's what I did:

For K₂:
If all concentrations are 1M, then Q should also be 1, right? If that is the case, then E= -0.224V - 0.0591/2*ln(1)=-0.224V. I then substituted this into nFE=RTln(K) to get (2 mol e)(96485 C/mol e)(-0.224V)=(8.31447 J/mol*K)(298 K)ln K, and then solved to get K₂=2.651*10^-8.

I did a similar process for the rest. What am I doing wrong?

[Borek]

Yes, standard potential is for 1M solutions, but it doesn't mean calculated K always equals 1.

Seems to me like your K₂ is reciprocal of the correct value.

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Yes, standard potential is for 1M solutions, but it doesn't mean calculated K always equals 1.

Did you mean that the Q isn't always 1? If that's true then, when is that the case?

I actually used the answer C (7.15*10⁵) and worked backwards to figure out that Q should've been -13.44, I think? (unless I messed up somewhere?), and I do know from before that Q₄ = Q₂/Q₃, and Q₂= [OH^-]², while Q₃=[NH₃]⁴/[Cu(NH₃)₄²⁺]. So if Q₂ and Q₃ aren't equal to 1, then how do I find them?

Sorry for not seeing the solution, but thank you so much so far! I really, really appreciate it!

[Borek]

Yes, standard potential is for 1M solutions, but it doesn't mean calculated K always equals 1.

Did you mean that the Q isn't always 1? If that's true then, when is that the case?

For the standard potential Q is 1, but when you are calculating K for any reaction, it can't be, as it would make no sense (K for every reaction being 1 sounds off, doesn't it?)

Trick is, standard potential is not at equilibrium. When Q is 1, system usually has a lot of free Gibbs energy that can be used. At equilibrium G is at minimum, and so effectively Nernst equation takes form 0=E₀-RT/nFln(K) (as we are at equilibrium it is K instead of Q now). That's where the nFE₀=RTln(K) comes from.

I actually used the answer C (7.15*10⁵) and worked backwards to figure out that Q should've been -13.44, I think?

Negative value for Q?

Do the simplest thing - use nFE=RTln(K) to calculate K₂ and K₃, then calculate their ratio.

If you are still wrong, most likely you messed up signs.