Impure copper containing Fe, Au and Ag as impurities is electrolytically refined. A current of 140 A for 482.5 s decreased the mass of the anode by 22.26 g and increased the mass of the cathode by 22.011g. Calculate the percentage of iron in impure copper.
(Given molar mass of Fe=55.5g `mol^(-1)`, molar mass of Cu=63.54 g `mol^(-1)`).

Explanation

In the purification of copper, impure copper is anode and pure copper is cathode. Increase in the mass of cathode is due to Cu deposited on it. Decrease in the mass of anode is due to `CutoCu^(2+)`(deposited on cathode), `FetoFe^(2+)` (remain dissolved int he solution) (Fe being more reactive than Cu), Ag and Au (being less reactive than Cu do not undergo oxidation) and fall below anode as anode mud.
Cu deposited on cathode (Pure Cu) can be calculated as follows:
`Cu^(2+)+2e^(-)toCu`
Quantity of electricity passed=`140xx482.5`coulombs=67550C ltbr `2xx96500C` deposit Cu=63.54g
`therefore67550C` will deposit Cu`=(63.54)/(2xx96500)xx67550g=22.239g`
Actual increase in the mass of cathode=22.011g
Cu not deposited due to Fe passing into the solution`=22.239-22011=0.228g`
Equivalent amount of Fe that passed into the solution`=(55.5)/(63.54)xx0.228g=0.199g`
Thus, Fe present as impurity=0.119g
Decrease in the mass of anode=Total impure `Cu(Cu+Fe+Au+Ag)=22.26g`
`therefore%` of Fe in impure `Cu=(0.199)/(22.26)xx100=0.89cong0.90`