cal. E.M.F

Asked by  | 26th Feb, 2008, 09:40: PM

Expert Answer:

Cd(s)ICd2+(.1m)  H+(.2m) I Pt2H2(.5))

Give E0 Cd2+I Cd = -.403V

It is a direct application of nernst equation, here for SHE E0 =0 but the conc = 1molar H+

so recheck and subsitute and calculate the answer.

Answered by  | 27th Feb, 2008, 04:27: PM

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