What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V?

Question

Given: A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25degrees C . The initial concentrations of Pb2+ and Cu2+ are 5.30×10−2 M and 1.60 M, respectively.

The Einitial cell that I calculated was 0.47. I know how to figure out the Q, but I don't know how to figure out the specific concentrations of Pb2+ and Cu2+ given only this information. Please help me! Thank you :)

Answer 1

Try this. Q = (Pb^2+)/(Cu^2+)= ?
You know Pb goes into soln so it will be 0.053+x and Cu^2+ comes out of soln and it will be 1.6-x.

Answer 2

It worked! Thank you so much :)

Answer 3

pb2+ +.....=pb