The +ve ions are discharged at the cathode. The -ve ions are discharged at anode.
Cathode:
#sf(K_((l))^(+)+erarrK_((l)))#
Anode:
#sf(2F_((l))^(-)rarrF_(2(g))+2e)#
The amount of charge in Coulombs which flows in 1 second is the electric current in Amps.
The total charge passed is given by:
#sf(Q=Ixxt)#
Converting hours into seconds we get:
#sf(Q=10.0xx2xx60xx60=7.20xx10^(4)color(white)(x)C)#
We now can find how much fluorine this will discharge.
From the anode 1/2 equation you can see that:
1 mole #sf(F_2)# needs 2 moles of electrons.
The charge carried by 1 mole of electrons = #sf(9.648xx10^(4)color(white)(x)C)#. This is known as the Faraday Constant.
So 1 mole of #sf(F_2)# needs #sf(2xx9.648xx10^(4)=19.296xx10^(4)color(white)(x)C)#
#sf(19.296xx10^(4)color(white)(x)C)# discharges 1 mole #sf(F_2)#
#:.##sf(1color(white)(x)C)# discharges #sf((1)/(19.296xx10^(4)))# moles #sf(F_2)#
#:.##sf(7.20xx10^(4)color(white)(x)C)# discharges #sf((1)/(19.296xxcancel(10^(4)))xx7.2xxcancel(10^(4)))# moles #sf(F_2)#
#sf(=0.37313)# moles #sf(F_2)#
I mole of fluorine has a volume of #sf(24.465color(white)(x)L)# under the conditions given.
#:.#0.3713 moles of fluorine has a volume of:
#sf(24.465xx0.37313=9.13color(white)(x)L)#
From the cathode 1/2 equation you can see that:
1 mole of K needs 1 mole of electrons.
So 1 mole of K needs #sf(9.648xx10^(4)color(white)(x)C)#
#sf(A_(r)[K]=39.098)#
#:.##sf(9.648xx10^(4)color(white)(x)C)# discharges #sf(39.098color(white)(x)g)# K
#sf(1color(white)(x)C)# discharges #sf((39.098)/(9.648xx10^(4))color(white)(x)g)# K
#:.##sf(7.20xx10^(4)color(white)(x)C)# discharges #sf((39.098)/(9.648xxcancel(10^(4)))xx7.2xxcancel(10^(4))color(white)(x)g)# K
#sf(=29.2color(white)(x)g)# K
