Lead ions are attracted to the cathode where they are discharged:
#sf(Pb^(2+)+2erarrPb)#
This means that 1 mole of #sf(Pb^(2+))# requires 2 moles of electrons to form 1 mole of #sf(Pb)#.
The charge on 1 mole of electrons is #sf(9.649xx10^(4))# Coulombs. This is termed 1 Faraday ( F ).
We can find the number of Faradays passed hence the moles and mass of lead discharged.
#sf(Q=It)#
#:.##sf(Q=1.57xx17xx60=1601.4color(white)(x)C)#
This is equal to #sf(1601.4/(9.649xx10^(4))=0.0166color(white)(x)F)#
#sf(2F)# will discharge #sf(1)# mole
#:.##sf(1F)# will discharge #sf(1/2)# mole
#:.##sf(0.0166F)# will discharge#sf(1/(2)xx0.0166=0.0083)# mole.
#sf(m=nxxA_r)#
#sf(m=0.0083xx207.2=1.72color(white)(x)g)#
This means the final mass of the elecrode will be:
#sf(7.63+1.72=9.35color(white)(x)g)#
