Impure copper containing Fe, Au, 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 cathode by ( 22.011 ~g ). Percentage of iron in impure copper is (Given molar mass ( Fe =55.5 ~g ~mol ⁻¹ ), molar mass ( Cu =63.54 ~g ~mol ⁻¹ ) )

Step-by-step solution

No. of gram equivalents of copper deposited = 22.011 31.77 =0.6298 No. of gram equivalents from the current (Q)=I t=140 482.5=67550 C No. of gram equivalents = 67550 96500 =0.7 Since only Cu and Fe are dissolved from the anode no. of gram equivalents of Fe =0.7-0.6928=0.0072 Therefore, mass of Fe =0.0072 27.75=0.1998 g % of Fe = Mass of Fe Mass of impurities 100= 0.1998 22.26 =0.89 or 0.89 %