# How do I calculate the cell potential?

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1. Jun 1, 2015

### Newton25

< 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.

Last edited by a moderator: Jun 1, 2015
2. Jun 1, 2015

### Staff: Mentor

Would it help if you ignore pH in the iron solution, and use it only for the hydrogen electrode? And yes, you should use pressure in Q, standard state calls for hydrogen pressure of 1 bar.

Actually question doesn't look good to me - if memory serves me well, solubility of Fe(OH)3 is so low Fe3+ can exist only in quite acidic solutions. pH 5 is way too high.

3. Jun 1, 2015

### Newton25

Yes, that would help. Thank you. I wasn't sure if I was able to ignore it or not.

So, would the Q then be equal to ([.00100]^4 * [antilog(-5.00)]^4)/ ([.50]^2 * [1.50]^4) ?

4. Jun 2, 2015

### Staff: Mentor

You are on the right track, but check if your reaction equation is balanced.