CBSE - XII Science - Chemistry - Electrochemistry
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.
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