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)`)

Options

A. 0.95
B. 0.85
C. 0.97
D. `0.90`

Answer

D

Explanation

`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%`