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Christophicus
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In an electrolysis experiment, a steady current of 0.40A is passed for 20 mins through a cell containing copper sulphate solution. Both electrodes in the cell are copper plates. Copper ions from the solution are attracted to the cathode and neutralised by gaining two electrons per ion. Each copper atom leaving the anode goes into solution as a copper ion by releasing two electrons. Calculate;
a) the total charge passed
b) the number of copper atoms deposited on the cathode
c) the mass deposited on the cathode given each copper atom has a mass of 1.1 x 10^-25kg

Answers given in book
a)480C
b)1.5 x 19^21
c) 0.165g
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BCHL85
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a) I = Q/t -> Q = It = 0.4*20*60 = 480C
b) Cu(+2) -2e -> Cu
If n is number of Cu(+2)-> Cu, so Q = n*2e
-> n = Q/2e =480/2*1.6E-19 = 1.5E21
c) m = n*1.1E-25 = 1.65E-4kg= 0.165g.
Is it right?
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BCHL85
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Looks like Chemistry question when write an equation like that :p:
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