Calculating E(cell) and Ratio of [Ni^2+]/[Zn^2+] in a Zn-Ni Cell

[requal]

Homework Statement

For a cell in which the reaction Zn(s)+ Ni^2+(aq)->Ni(s)+ Zn^2+(aq) calculate the E(cell) when [Ni^2+]=0.05M, [Zn^2+]=0.85M.

Then, find the ratio of [Ni^2+]/[Zn^2+], when the cell is "flat"

Homework Equations

The simplified nernst equation ; E(cell)=E*(cell)-(0.0592V/n)xlogQ, log=log base 10

The Attempt at a Solution

Well I worked out the standard potential by adding half equations

Ecell=ENi-Ezn=0.51V

than because two electons are transferred, n in the nernst equation is 2.

Ecell= 0.51V-0.0592V/2log[0.85/0.05]

I plug that into my calculator, and I get 0.47V; the answer is 0.49V?
I tried doing the second part, and I got it wrong obviously, because I got the first part wrong

 

[Borek]

I got 0.48V (0.477), it probably depends on values of standard potentials used and/or rounding errors.

But I don't see how the second part depends on the first, looks to me like these are completely separate problems.