If 3.25 x 10^-3 kg of gold is deposited on the negativeelectrode of an electroly

Question

If 3.25 x 10^-3 kg of gold is deposited on the negativeelectrode of an electrolytic cell in a period of 2.78 h, what isthe current through the cell in this period? Assume that thegold ions carry one elementary unit of positive charge. Hint given by teacher?? kg to mol to Q to I, M (au) = 197g/mol, density (au) is 19,300 kg/m^3 Thanks a lot! If 3.25 x 10^-3 kg of gold is deposited on the negativeelectrode of an electrolytic cell in a period of 2.78 h, what isthe current through the cell in this period? Assume that thegold ions carry one elementary unit of positive charge. Hint given by teacher?? kg to mol to Q to I, M (au) = 197g/mol, density (au) is 19,300 kg/m^3 Thanks a lot!

Explanation / Answer

1 F of charge is required to deposit 1 mol of gold

mol of gold= 3.25 x 10^-3 kg/0.197 = 0.0165 mol

so charge = 96500*0.0165= 1592.25 C

time= 10008 seconds

so,current I=Q/t= 0.159A

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