Consider the cell: (Pt) H2/H+ || (Pt) H+/H2. In the anode half-cell, 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 half-cell is identical to the first except that the solution around the platinum electrode has a pH of 0.0. What is the cell voltage?

Question

Consider the cell: (Pt) H2/H+ || (Pt) H+/H2. In the anode half-cell,  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 half-cell is identical to the first except that the solution around the platinum electrode has a pH of 0.0. What is the cell voltage?

DrBob222 · Answer

For the anode written as a reduction:
E = Eo - (0.0592/n)log(pH2/H^+)
Eo = 0, and H^+ = 1E-7, 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.

Anonymous · Answer

nevermind i got it.. the answer is -.414 thanks though

DrBob222 · Answer

right. I dropped the - sign when I typed it in.