< Mentor Note -- thread moved to HH from the technical chemistryforum, so no HH Template is shown > I was given this problem: A voltaic sell uses the reaction 4Fe3+ + H2 --> 2Fe2+ +2H+. Find the cell potential when [Fe3+]=1.50 M, PH2=.50 atm, and [Fe2+]=.00100 M and the pH of both anode and cathode compartments of the cell=5.00. The reduction potentials are Fe3+ +1e --> Fe2+ E=.771V H2--> H+ + 2e E=0 V I'm having quite a bit of difficulty with it. My first step was to balance the reaction equation and I split it into the half reactions. For the oxidation half, I got 2H2--> 4H+ + 4e. For the reduction, I got 4Fe3+ +4e --> 4Fe2+. And my E initial was .771 V. My next step was to use the Nernst equation. I had E=.771 - ((.0592)/4)logQ. Solving Q is the part that's giving me problems. First, I wasn't sure if I could use the pressure of H2 as everything else was a concentration. Second, the pH of both compartments being 5.00 is messing me up. If it was just the anodic compartment, I would use the antilog to determine the concentration of hydrogen ion. However, since it's 5.00 in both compartments, I have no clue what to do with that. If anybody could help with anything, it'd be greatly appreciated! Thanks.