chemistry
posted by Anonymous .
Consider the cell: (Pt) H2/H+  (Pt) H+/H2. In the anode halfcell, hydrogen gas at 1.0 atm is bubbled over a platinum electrode dipping into a solution that has a pH of 7.0. The other halfcell is identical to the first except that the solution around the platinum electrode has a pH of 0.0. What is the cell voltage?

For the anode written as a reduction:
E = Eo  (0.0592/n)log(pH2/H^+)
Eo = 0, and H^+ = 1E7, E then written as an oxidation will be the negative of that.
E for the cathode written as a reduction:
same equation with different numbers.
Then E cell = Eoxdn + Eredn
I think the answer is approximately 0.50v. 
nevermind i got it.. the answer is .414
thanks though 
right. I dropped the  sign when I typed it in.