Question

Impure copper containing Fe, Au, Ag as impurities is electrolytically refined. A current of 140A for 482.5s decreased the mass of the anode by 22.26g and increased the mass of the cathode by 22.011g. Percentage of iron in impure copper is:
(Given molar mass of Fe=55.5gmol-1, molar mass of Cu=63.54gmol-1)
A) 0.85
B) 0.90
C) 0.95
D) 0.97

Answer 1

Hint : As stated by Faraday’s second law, mass of any substance liberated or deposited at the anode or cathode is directly proportional to their equivalent weight if the same amount of current is passed through different electrolytes or elements connected in series.
Formulas used:
-Equivalent weight: Equivalent weight the amount of any substance that is required to react with a fixed amount of another substance.
$Eq.wt. = Mol.wt./Nfactor$ (1)
Where, Eq.wt. =Equivalent weight
Mol.wt. =Molecular weight
N-factor=Number of electrons lost or gained or the change in oxidation
state
-Faraday’s Second Law: Faraday’s second law of electrolysis states that when the same amount of electricity is passed, through different electrolytes, the masses of different ions that are liberated at the electrodes are directly proportional to their equivalent weights.
${W_1}/{W_2} = {E_1}/{E_2}$ (2)
-Current is the rate of flow of charge.
$I = Q/t$
So, $Q = It$ (3)
Where, Q= charge
I=current
t=time

Complete step by step solution : In purification of Cu, anode is the impure Cu and cathode is pure Cu. Increase in the mass of cathode is due to the deposition of Cu on it. Mass of anode is decreased due to Cu deposition on cathode, Fe oxidation (due to it being more reactive than Cu), Ag and Au (being less reactive than Cu) falling down as anode mud.
-It is given that: I=140A, t=482.5s. Using this data we will first calculate the amount of copper that should be deposited in the given time:
So first, use equation (3) to calculate the amount of charge passed in t time:
$Q = It$
Q=140×482.5 =67500 C
Now, since we know that the amount of element deposited for a charge of 1F is equal to its equivalent weight. We calculate it by using equation (1):
Cu deposited for a charge of 1F = Eq.wt. of Cu
$Eq.wt.(Cu) = Mol.wt.(Cu)/Nfactor(Cu)$
= 63.54/2
= 31.77g
Since, 1 Faraday charge=96500 C, we use the unitary method to calculate the amount of pure copper that should be deposited on the cathode when the charge is 67550 C:
Amount of pure copper to be deposited= (31.77/ 96500) ×67550
= 22.239 g pure Cu (4)
- It is given that increased mass of cathode (Cu) = 22.011 g (5)
-From (4) and (5) we see that the amount of copper that should have been deposited on the cathode is more than the actual increased mass of the cathode. So we calculate the amount of copper not deposited due to impurities or due to Fe passing into the solution:
Impurities = amount of copper to be deposited – increased mass of cathode
= 22.239 – 22.011
= 0.228 g
-Now we need to calculate the total amount of iron (Fe) present in the impure metal. For this we use Faraday’s second law. Using equation (2):
${W_1}/{W_2} = {E_1}/{E_2}$
Where, W1=weight of Fe that passed into solution (to be calculated)
W2=weight of Cu that passed into solution =0.228 g
E1=equivalent weight of Fe = 27.55 g eq-1
E2=equivalent weight of Cu = 31.77 g eq-1
W1/0.288 = 27.55/31.77
W1 = (27.55 × 0.228) / 31.77
= 0.199 g
-Fe present as impurities= 0.199 g
-Decrease in the mass of anode = weight of impure metal= 22.26 g
So, percentage of pure Fe in impure metal:
= (wt. of Fe present/wt of impure metal) × 100
= (0.199 / 22.26) × 100
= 0.89 which is approximately = 0.90 %

So, the correct option is: B) 0.90

Note : Always solve such questions step by step to avoid any problems. Never confuse yourself between cathode (negatively charged electrode) and anode (positively charged electrode). Be very careful while putting the values in the formulas.