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Newton25
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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.
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.
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