[PhilMaq]

I'm having trouble on how to figure out the last part of one of my hw questions.
Here's the given info. A voltaic cell has standard cell potential (E_cell) of 0.53 V.
Here's the balanced redox reaction : Zn(s) + Ni^2+(aq)-->Zn^2+(aq) + Ni(s)

Question: What are the concentrations of Ni^2+ and Zn^2+ when the cell potential drops to 0.45V?

I'm using this equation Ecell=(Ecell at standard conditions) - [(0.0592/n)(logQ)]
so... 0.45=0.53- [(0.0592/2)(logQ)]
(0.0296)(logQ)=.08
logQ=2.7
Q=501.19

so the concentration of the Zn2+ over Ni2+ is equal to 501.19....how do I figure out the concentration of each?
I have the answer btw, I don't need that. I want an explanation on how to do it. The answer is [Ni2+]=0.003M and [Zn2+]=1.60M

[fredp]

I think you can form another equation by the stoichiometry.  You know the relative amount of each of the reactants consumed.  This, combined with ratio you already calculated, should lead to the answer. 

[FreeTheBee]

You can interpret the question as: Assume the cell is built as a standard cell, so with unit concentrations/activities. Then, let it run until the potential drops to 0.45V. If you read it like this, you know the starting conditions, so you can calculate the final concentrations. I don't get the answer you gave though, the nickel concentration I get is reasonably close to the answer you posted, but the zinc concentration seems wrong to me.

[fairlady95]

First, solve for Q using the Nernst equation. Knowing the formula:
Zn(s) + Ni^2+(aq)-->Zn^2+(aq) + Ni(s)

You know Q=[Zn2+]/[Ni2+]

The initial concentration of Zn2+ is 1.0M and the initial concentration of Ni2+ is 1.5M.
Using an ICE table,we obtain the formula for Q:

Q=(1.0+x)/(1.5-x)

Solve for x and then substitute the values in. I agree, the concentration of Zn in the answer is wrong.