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Topic: Using the Nernst equation & calculating new concentration  (Read 7991 times)

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Offline angelxstyl

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Using the Nernst equation & calculating new concentration
« on: July 30, 2009, 05:16:50 AM »
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 Ecell = E°cell - (0.0592/n)log([Cu2+]/[Ag+]2)
and was told that n = 2.

Here's what I did...

Ecell = E°cell - (0.0592/n)log([Cu2+]/[Ag+]2)
0.40 = E°cell - 0.0296log(0.1/0.01)
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?

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