Question

Calculate emf of the following cell
Cd/\(Cd^{2+}\) (.10 M)//\(H_+\) (.20 M)/\(H_2\) (0.5 atm)/Pt
[Given E° for \(Cd^{2+}\) /Cd = -0.403V]
\(E_{cell}\) = \(E°_{cell}\) – 0.0591/n log [\(Cd^{2+}\)] \([\\H^+\\]^2\)
(E°_{cell}\) = 0 – (–.403V) =0.403V
=0.0403 – 0.0591/2 Log (0.10) X 0.5/(0.2)2= 0.400V

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