Help with Nernst, please??

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Aug 12, 2005
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So here I am, freshman in Pharmacy. Problem is, it's been 4 years since I've done any chemistry and boy am I rusty.

I have to have this lab report in by monday. It's one of the trusty Nernst equation thingamajigs. It would appear that the E^o is 0.5387. There were two electrochemical measurements and it turned out as follows:

initially for experiment 1:

Ecell(V) is 0.7996
[Ag+]i is 0.01
[NH3]i is 0.08
[Ag+]f is 1.59186 mol dm-3
[NH3]f is 0.08 mol dm-3
[complex] is 0.01 mol dm-3
Kc is 0.78524493

Experiment 2:
Ecell (V) is 0.1069
[Ag+]i is 0.01
[NH3]i is 0.04
[Ag+]f is 0.0305991
[NH3]f is 0.4
[complex] is 0.01 mol dm-3
Kc is 0.8170

Average value of Kc is 0.44770993 (which I'm fairly sure is correct??)

So, here's the question:

The expression for the equilibrium constant Kc=[Ag(NH3)n+]/([Ag+][NH3]fn) allows you to write two simultaneous equations in two unknowns (n and Kc) from your results. Can you use your results to show that n=2?


If Kc=Ag(NH3)n+]/([Ag+][NH3]fn), then

for (Exp.1) Kc= [0.01]/[0.1273488n]
for (Exp.2) Kc= [0.01]/0.01223964n]

so, for Exp.1, 0.078524493/n=0.078524493 (the Kc that I found above for experiment 1), therefore n=0.999745, n~1

Exp2, 0.8170/0.81701=n, therefore n=1

[n1]&[n2}=2, therefore n=2

I know the final line doesn't make any sense, but it's the only way I could twist it so that n=2! I feel so close, yet so far. Help, please?