Question

To electrodeposit all the Cu and Cd from a solution of CuSO4 and CdSO4 required 1.20 Fof electricity (1F=1mole−). The mixture of Cu and Cd that was deposited had a mass of 50.42 g . What mass of CuSO4 was present in the original mixture?

Answer 1

Cu2+ + 2e ------> Cu(s)

Cd2+ + 2 e -------> Cd (s)

let x = moles of Cu and y = moles of Cd

x + y = 1.2 F/2 = 0.6

x = 0.6 - y ------- (1)

and x (63.5 ) + y (112.4) = 50.42

Substitute (1) in the above equation,

(0.6 - y) (63.5) + y (112.4) = 50.42

38.1 - 63.5 y + 112.4 y = 50.42

48.9 y = 12.32

y = 0.252 mol of Cd

Then x = 0.6 - 0.252 = 0.348 mol of Cu

And 1 mol of Cu = 1 mol of CuSO4 = 63.5 + 32 + 4(16) = 159.5 g. of CuSO4 is present the original mixture.