[UG]

Hi everyone,
Can someone please check if I've done this problem correctly?
It goes:
The addition of aqueous ammonia to the cathode compartment of the cell described below reduces the cell potential. Likewise, addition of aqueous ammonia to the anode compartment increases the cell potential.
Zn(s) / Zn²⁺ (aq) // Cu²⁺ (aq) / Cu(s)
Assume that [Cu²⁺]_original = [Zn²⁺]_original = 0.10 M and that both form stable complexes with ammonia (overall formation constants >10¹⁰).
Show how to estimate the ratio of the overall formation constant for [Cu(NH₃)₂]²⁺ to the overall formation constant for [Zn(NH₃)₂]²⁺ if the cell potential is observed to be 0.96 V after adding equal amounts (assume a large excess) of 6.0 M NH₃ to each compartment (assume that T=298K)

Okay, here's my way, and I am very sceptical about it  

E^o (Zn/Zn²⁺) = -0.76 V
E^o (Cu/Cu²⁺) = 0.34 V

Using the form of the Nernst equation valid at 298 K

E = E^o - [0.0257/n In(Q)]   n=2 mol electrons

So for zinc, E = -0.76 -0.01285 In(Q₁)
And for copper, E = 0.34 - 0.01285 In(Q₂)

Since E_cell = E(Cu/Cu²⁺) - E(Zn/Zn²⁺)
         0.96 = [0.34 -0.01285 In(Q₂)] - [-0.76 -0.01285 In(Q₁)]
         0.96 = 1.1 - 0.01285 In(Q₂) + 0.01285 In(Q₁)
        -0.14 =  - 0.01285 In(Q₂) + 0.01285 In(Q₁)
Divide through by -0.01285
10.9 = In(Q₂) - In(Q₁)
       = In(Q₂/Q₁)
e^10.9 = Q₂/Q₁
       = 54000 ?

The overall formation constant for [Cu(NH₃)₂]²⁺ is 54000 times larger than the formation constant for [Zn(NH₃)₂]²⁺??  
Is my way even remotely correct?  

[Borek]

Q₂/Q₁ = 54000

What are Q₂ and Q₁?

[UG]

The reaction quotients

Zn²⁺ (aq) + 4NH₃ (aq)  Zn(NH₃)₄²⁺ (aq)    Q₁
Cu²⁺ (aq) + 4NH₃ (aq) Cu(NH₃)₄²⁺ (aq)     Q₂

[Borek]

No, you have not used standard potentials for these reactions.

[UG]

Right, so the non-standard values can be calculated by:

E = E^o - [0.0257/n In(Q)] ?

[Borek]

E = E0 + RT/nF ln[Meⁿ⁺] will do - just think what [Meⁿ⁺] means. Can you calculate it as a function of K_f?

[UG]

Does Meⁿ⁺ mean the Zn²⁺ and Cu²⁺ ions?

And if we take Zn²⁺

K_f[(Zn(NH₃)₄²⁺ (aq))] =  [Zn(NH₃)₄²⁺]/[Zn²⁺][NH₃]⁴

Rearrange to get [Zn²⁺] or Meⁿ⁺ as = [(Zn(NH₃)₄²⁺]/K_f [NH₃]⁴ ?


-----
Btw, is this a typo? E = E⁰ + RT/nF ln[Meⁿ⁺]
should it not be E = E⁰ - RT/nF ln[Meⁿ⁺]

[Borek]

Does Meⁿ⁺ mean the Zn²⁺ and Cu²⁺ ions?

Me like Metal, so it stands for Zn, Cu or any other.

Rearrange to get [Zn²⁺] or Meⁿ⁺ as = [(Zn(NH₃)₄²⁺]/K_f [NH₃]⁴ ?

Can you tell what are approximate concentrations of complex and free ammonia (based on the information given in the question)? Can you put this calculated value into Nernst equation?

Btw, is this a typo? E = E⁰ + RT/nF ln[Meⁿ⁺]
should it not be E = E⁰ - RT/nF ln[Meⁿ⁺]

No, that's correct. General form of the Nernst equation is

E = E0 + RT/nF ln [Ox]/[Rеd]

[Ox]/[Rеd] can be replaced by reaction quotient, just remember to have oxidized side in the nominator. If you put it in denominator, you should flip the sign.

[UG]

Can you tell what are approximate concentrations of complex and free ammonia (based on the information given in the question)? Can you put this calculated value into Nernst equation?

Ahhh, okay, I am starting to 'see' this problem a bit better
So for the zinc:

[Zn²⁺] = [~0.10]/ K_f1 [6.0]⁴

And for copper

[Cu²⁺] = [~0.10]/ K_f2 [6.0]⁴

No, that's correct. General form of the Nernst equation is

E = E0 + RT/nF ln [Ox]/[Rеd]

[Ox]/[Rеd] can be replaced by reaction quotient, just remember to have oxidized side in the nominator. If you put it in denominator, you should flip the sign.

I see, so if I decide to stick with the '-' sign it's going to be E = E⁰ - RT/nF ln[1/Meⁿ⁺] And everything will be flipped around. That makes sense too.

So now, I go:

E(Zn²⁺/Zn) = E⁰ - RT/nF ln (K_f1 [6.0]⁴/[~0.10])

and E(Cu²⁺/Cu) = E⁰ - RT/nF ln (K_f2 [6.0]⁴/[~0.10])

0.96 V = [0.34 - 0.01284 ln (K_f2 [6.0]⁴/[~0.10])] + [0.76 + 0.01284 ln (K_f1 [6.0]⁴/[~0.10])]
0.96 = 1.1 - 0.01284 In (12960 K_f2) + 0.01284 In (12960 K_f1)
10.90 = In (12960 K_f2) - In (12960 K_f1)
10.90 = In (12960 K_f2)/(12960 K_f1)
54176 = 12960 K_f2/12960 K_f1
K_f2/K_f1 = 4.18


Is this the correct method?

[Borek]

I have not checked the numbers, but seems to me like you got the idea right. Just don't write In when you mean ln, it is natural logarithm, not indium

[UG]

Hey, could someone help clear up some things about pH meters for me, I've read many different sites but apparently I'm still a bit hazy on this...and they never seem to tell you how all of this ties in with the Nernst equation  One question I have which goes like this:
The glass electrode in pH meters acts as acid-base material due to the charged oxygens at the surface of the glass. The reaction goes:
MemH⁺(s) + H₂O  Mem(s) + H₃O⁺
pH meters use a combination of the glass electrode and a silver/silver chloride electrode. When the two electrodes were immersed in a buffer of pH=7.00, the potential is 0.433 V. When put in a second solution, the potential is 0.620 V. Calculate the pH.

So the first thing I'm a bit confused about is the reaction: MemH⁺(s) + H₂O  Mem(s) + H₃O⁺ shouldn't the reaction be the other way round? And this isn't a half reaction correct? There is no transfer of electrons, it's just acid + base reaction?
Secondly, the potential that they are talking about, that is a measure of the difference in potential between [H⁺] on the glass surface and [H⁺] inside the electrode against the known potential of the Ag/AgCl electrode? Have I got this part correct?
Finally, when I am calculating the pH, do I assume that the Ag/AgCl electrode is a reference electrode like the standard hydrogen electrode? So it has potential = 0 V?
I understand the Nernst equation should be E = E⁰  - RT/zF ln(1/[H⁺])
and that 'z' = 1 electron. Where did this come from? Where is the exchange of electrons occuring?

[UG]

Hi everyone, I'm a bit stuck on another electrochemistry problem, it asks:
Calculate the solubility product of CuI, given:
Cu⁺ + e^-  Cu             E⁰ = 0.52 V
Cu²⁺ + e^-  Cu⁺          E⁰= 0.15 V
Cu²⁺ + I^- + e^-  CuI     E⁰ = 0.86 V
I₂ + 2e^-  2I^-              E⁰ = 0.54 V

I am not sure how this all fits with the Nernst equation but here's what I did
So I am trying to find K_sp for CuI  Cu⁺ + I^-
So I reasoned that any Cu⁺ formed will undergo a disporportionation reaction 2Cu⁺ (aq)  Cu(s) + Cu²⁺ (aq) for which the calculated E⁰cell = 0.37 V
Then the Cu²⁺ formed will react with I^- and reduction occurs: Cu²⁺ + I^- + e^-  CuI
While Cu is oxidised: Cu Cu⁺ + e^-
Overall E⁰cell for the reaction Cu²⁺ + I^- + Cu  Cu⁺ + CuI = 0.34 V
So then do I combine the two equations? Like
2Cu⁺ (aq)  Cu(s) + Cu²⁺ (aq)        E⁰cell = 0.37 V
Cu²⁺ + I^- + Cu  Cu⁺ + CuI           E⁰cell = 0.34 V

After cancelling out some species, I get: Cu⁺ + I^-  CuI
So now, do I add up the voltages, like 0.37 V + 0.34 V = 0.71 V ? Is this the E⁰cell for the reaction Cu⁺ + I^-  CuI? So then I put this into the Nernst equation:
E⁰ = RT/nF ln(1/[Cu⁺][I^-])
0.71V / (8.314 x 273/96485) = ln(1/[Cu⁺][I^-])
30.18 = ln(1/[Cu⁺][I^-])
1.28 x 10¹³ = 1/[Cu⁺][I^-]
So K_sp = 7.80 x 10^-14

Is my method correct?