Here the amount of electricity passed through the electrode in 16minutes or 960s @ 2mA is =2×960=1920 C.
For the passage of 96500 C of electricity the number of moles of Cu2+ ion deposited is 1 gm equivalent or 1/2 mole.
So for the passage of 1920 C electricity the Cu2+ ion deposited will be
1/2×1920/96500= 0.01 mole.(approx)
As the passage of electricity reduces the absorbance by 50% which is proportional to concentration, we can say that number of moles of Cu2+ ions in original 250 mL solution will be =0.02 moles.
Hence concentration of original solution was
=0.02/250×1000=0.08M
