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The cell Cu/cu(No3)2(100ml,1M)||AgNo3(100ml,1M)/Ag was used as an electrolytic cell.a current of 0.96 A was passed for 5 hr and there after cell was allowed to function as galvanic cell assuming electrode reaction involved only Cu/Cu2+ and Ag/Ag+. find the change in emf of the cell.given (E0 cu/cu2+=-3.21ev,E0ag/ag+=-0.8ev).

shubham garg , 11 Years ago
Grade 12
anser 1 Answers
Pankaj

Last Activity: 10 Years ago

The emf of cell can be calculated using formula
E= E^0_c_e_l_l-\frac{0.0591}{n}log\frac{[Cu^2^+]}{[Ag^2^+]}
Where [Cu2+] =[Ag2+] = 1 M
n = 2
and
E0cell = Eoxid – Ered = -3.21 + 0.8 = -2.41

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