Question #24d99

The first step is to find the quantity of electricity, in Coulombs, to discharge #1.63"g"#.

Once we know this we can find the time this will take since we are told the current, which is the rate of flow of charge.

Nickel(II) ions are discharged:

#Ni_((aq))^(2+)+2erarrNi_((s))#

#:.# 1 mole #"Ni"^(2+)# needs 2 moles of electrons to produce 1 mole #"Ni"#.

We can find the charge on 1 mole of electrons by multiplying the electronic charge by the Avogadro Constant. This gives #9.64xx10^(4)"C"# and is referred to as the Faraday Constant #"F"#.

#:. 58.7"gNi"^(2+) # require #2"F"# to produce #58.7"gNi"#

#2"F"=2xx9.64xx10^(4)=1.928xx10^(5)"C"#

So:

#58.7"g Ni"# is formed from #1.928xx10^(5)" ""C"#

#:.1"gNi"# is formed from #(1.928xx10^(5))/(58.7)" ""C"#

#:.1.63"gNi"# is formed from #(1.928xx10^(5))/(58.7)xx1.63" ""C"#

#=5.35xx10^(3)" ""C"#

Electric current is the rate of flow of charge:

#I=Q/t#

#:.t=Q/I#

#t=(5.35xx10^(3))/13.7=390.5"s"#

#=390.5/60=6.5"min"#

In reality a current of this size, as well as being very dangerous, would not give a very good result.

The quality of the deposit at the cathode is favoured by a low current density which is current/area. So a low current over a high area gives a deposit that binds well to the surface.

A high current density like this would give a very spongy deposit which would fall off the cathode very easily.