[sammmyx_]

Okay, so we did an experiment in our Chem lab where we made a galvanic cell with [Cu(NH₃)₄]²⁺ on one side and a copper reference electrode on the other.

We had to make the  [Cu(NH₃)₄]²⁺ by adding 8mL of 0.2M CuSO4 and 8mL of 1.0M NH3 to each other to make the solution.

We tested the galvanic cell and it gave us a reading of 0.134V.

Cu | [Cu(NH₃)₄]²⁺ (aq), NH₃ (aq) || Cu²⁺ (aq) | Cu

We had to use this cell voltage to find the value of the stability constant, K_stab

The hint for the questions are:

1.Calculate the concentrations of  [Cu(NH₃)₄]²⁺ (aq) and NH₃ (aq) in the expression for Q from the initial concentrations of Cu²⁺ and NH₃ (after allowing for mutual dilution) assuming this reaction goes to completion: Cu²⁺ (aq) + 4NH₃    [Cu(NH₃)₄]²⁺ (aq).

2. The concentration of Cu²⁺ in the expression for Q is the concentration in the reference electrode.

Question 1:

Calculate the concentrations of NH₃ and [Cu(NH₃)₄]²⁺ in the left-hand cell. These are the concentrations referred to as × and γ respectively in: Cu²⁺ (aq, 1.0 mol/L) + 4NH₃ (aq, × mol/L)  [Cu(NH₃)₄]²⁺ (aq, γ mol/L). Do not forget to allow for the mutual dilution which happens when you mix solutions.

Question 2:

Use the concentrations from step 1 and the known concentration of Cu²⁺ (aq) in the reference electrode to calculate Q.
Use the formula:

Q = [Cu(NH₃)₄]²⁺ / [Cu²⁺] [NH₃]⁴.

Question 3:
Use the re-arranged form of the Nersnt equation given to calculate ln K^stab and hence K^stab from the measured cell potential and the value of Q calculated above.

ln K = nFE/RT + ln Q

Thanks!

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

You are very precisely told what to do. So, what you did?

[sammmyx_]

Oh! My bad! Here's what I did 

Q1) Cu²⁺ (aq, 1.0 mol/L) + 4NH₃ (aq, × mol/L)  [Cu(NH₃)₄]²⁺ (aq, γ mol/L)

From initial:

n(NH₃) = 0.008L × 1.0M = 0.008 moles
               γ = 1.0M right? 
n(Cu²⁺) = 0.008L × 0.02M = 0.0016 moles, therefore it is the limiting reactant and that would mean that n{[Cu(NH₃)₄]²⁺} = 0.0016 moles @ 0.016L, so the concentration of × = 0.1M

Q2)
Q = [Cu(NH₃)₄]²⁺ / [Cu2+] [NH₃]⁴

Q = [0.1] / [1][1]⁴
Q = 0.1

Q3)
ln K = nFE/RT + ln Q

ln K = 2×96485×0.134 / 8.314×298 + ln(0.1)
ln K = 8.134
K_stab = e^8.134
K_stab = 3409

I'm really not sure about my answer to the first question. It seems kinda fuzzy.

[mjc123]

Your calculation of [NH₃] is wrong. You've worked out the number of moles of Cu(NH₃)₄²⁺. Each mole of this species takes up 4 moles of NH₃. So how many moles of free NH₃ are left? And what is the volume of the solution?

[sammmyx_]

Oh! My bad! Here's what I did 

Q1) Cu²⁺ (aq, 1.0 mol/L) + 4NH₃ (aq, × mol/L)  [Cu(NH₃)₄]²⁺ (aq, γ mol/L)


Initial moles of NH₃ = 0.008L×1.0M = 0.008 moles
Initial moles of CuSO₄ = 0.008L×0.2M = 0.0016 moles, limiting reactant as 0.0016 moles of Cu²⁺ × 4 moles of NH₃/1 mole of Cu²⁺ = 0.0064 moles NH₃.

n{[Cu(NH₃)₄]²⁺} = 0.0016 Cu²⁺ × [Cu(NH₃)₄]²⁺ = 0.0016 moles

n(NH₃) that do not react = 0.0080 - 0.0064 = 0.0016 moles.

Final volume = 16mL = 0.016L

[[Cu(NH₃)₄]²⁺] = 0.0016 moles / 0.016L = 0.1M = y
[NH₃] = 0.0016 moles / 0.016L = 0.1M = x


Q2)
Q = [Cu(NH₃)₄]²⁺ / [Cu²⁺] [NH₃]⁴

Q = [0.1] / [1][0.1]4
Q = 1000


Q3)
ln K = nFE/RT + ln Q

ln K = 2×96485×0.134 / 8.314×298 + ln(1000)
ln K = 17.35
Kstab = e^17.35
Kstab = 34092163 or 3.41×10⁷

I think I've got it