A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 C. The initial concentrations of Pb+2 and Cu+2 are 0.0500 M and 1.50 M, respectively.

Question

A. What is the initial cell potential?
B. What is the cell potential when the concentration of Cu+2 has fallen to 0.200 M?
C. What are the concentrations of Pb+2 and Cu+2 when the cell potential falls to 0.35 V?

I answered A and B. For A I got: 0.43 V and for B I got 0.46 V. I cannot figure out part C though. I tried 0.35V=O.47V-((0.0592/2)*log(0.0500+X/1.50-X))but the answer doesn't seem right. I found X to equal 1. When I plug it all back in, it doesn't work though. Is my equation wrong? So confused...