Impure copper containing Fe, Au, Ag as impurities is electrolytically refined. A current of 140 A for482.5 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))

Impure copper containing Fe, Au, Ag as impurities is electrolytically

1.	Impure copper containing Fe, Au, Ag as impurities is electrolytically refined. A current of 140 A for482.5 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))
Answer» 0.95 0.85 0.97 `0.90` Solution :`Cu^(2+)+2e^(-)RARR Cu`, EQUIVALENT wt. `=(63.54)/(2)=31.77` `Fe^(2+)+2e^(-)rarr Fe`, equivalent wt.`=(55.5)/(2)=27.75` Mass inceased at cathode is due to deposition of Cu. Hence, no. of gram equivalents of Cu deposited `=(22.011)/(31.77)=0.6928` Now, using Faraday's FIRST law, `Q=it = 140 A xx 482.2 s = 67550C` `96,500C-31.77 g Cu` `67550 C - ?` `(31.77)/(96, 500)xx(67550)/(1)=22.239` g of pure Cu deposited But, according to the question, the mass of cathode only increases `= 22.011g` Hence, `22.239-22.011=0.228g` Now, `31.77g Cu -= 0.228` g deposited `27.75 g Fe -= (0.228)/(31.77)xx27.75=0.199` g Fe deposited % of Fe `= ("Mass of Fe")/("Mass of impurities (at anode)")xx100` `= (0.199)/(22.26)xx100=0.894~~0.90%`