[angelxstyl]

I'm having some trouble with calculations...

Initially, I had two beakers, one with 0.1M CuSO₄ solution and the other with 0.1M AgNO₃ solution. I measured the potential and it was 0.40V. Then, I added 10mL of 6M NH₃ to the CuSO₄ and measured the potential again.. which came out to be 0.84V

How do I calculate the concentration of that free Cu(II) ion that is in equilibrium with the complexed Cu(II) ion, Cu(NH₃)₄²⁺ in the solution?

The equation that I was told to use was E_cell = E°_cell - (0.0592/n)log([Cu²⁺]/[Ag⁺]²)
and was told that n = 2.

Here's what I did...

E_cell = E°_cell - (0.0592/n)log([Cu²⁺]/[Ag⁺]²)
0.40 = E°_cell - 0.0296log(0.1/0.01)
E°_cell = 0.4296

0.84 = 0.4296 - 0.0296log([Cu2+]/0.01)
0.4104 = -0.0296log([Cu²⁺]/0.01)
10^-13.86 = [Cu²⁺]/0.01
1.38 x 10^-14 = [Cu2+]/0.01
[Cu²⁺] = 1.38 x 10^-16 M

...but I don't know if that's the right answer?