Posted by Anonymous on Tuesday, March 27, 2012 at 7:43pm.
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.
nevermind i got it.. the answer is -.414
thanks though
right. I dropped the - sign when I typed it in.
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