Question

Assume that impure copper consists of iron, gold, silver impurities. After passing current of $140A$ for $482.5\sec $ , the mass of anode decreased by $22.260g$ and the cathode increased in the mass by $22.011g$ . Estimate the percentage of iron and copper originally.

Answer 1

Hint:Electrolysis is defined as the decomposition of a substance by an electric current. There is a chemical change in which one gains electrons and one loses electrons. Coulomb is the unit of charge.

Complete step by step answer:
REACTION:
A.At cathode: $C{u^{2 + }} + 2{e^ - } \to Cu$ (reduction)
At anode: $Fe \to F{e^{2 + }} + 2{e^ - }$ (oxidation)
Mass deposited at cathode $ = 22.011g$
Mass of anode $ = 22.260g$
From this we get to know the mass of impurities:
Mass of impurities $ = $ mass of anode $ - $ mass of cathode
Substituting the values we get,
Mass of impurities $ = 22.260 - 22.011$
Mass of impurities $ = 0.249g$ .
Here the impurities are iron, gold and silver which do not take part in the reaction and settle downs below anode.
B.we will calculate the charge using the formula: $Q = i.t$
Where, $i = $ current, $t = $ time
given data: $i = 140A$
$t = 482.5s$
$\therefore Q = i.t$
Substituting the values we get,
$Q = 140 \times 482.5$
$Q = 67550C$ .
We will convert the value of charge from coulombs to faraday $\left( F \right)$ using this formula: $F = \left( {\dfrac{C}{{96500}}} \right)$
Substituting the value of coulombs we get:
$F = \dfrac{{67550}}{{96500}}$
$F = 0.7$
C.Now we will calculate charge for iron and copper respectively using the formula: $F = \dfrac{m}{{MW}} \times n$
Where, $F = $ faraday
$m = $ mass
$MW = $ molecular weight
$n = $ number of electrons.
Given data:
Mass of copper at cathode $ = 22.011g$
Molecular weight of copper $ = 63.5$
i.Charge of copper
$F = \dfrac{m}{{MW}} \times n$
Substituting the values we get,
$F = \dfrac{{22.011}}{{63.5}} \times 2$
$F = 0.693$
ii.Charge of iron:
Total charge $ - $ charge of copper
$ = 0.7 - 0.693$
$ = 0.007$
Using the value of charge we can find the mass of iron
$F = \dfrac{m}{{MW}} \times n$
Molecular weight of iron $ = 56$
Number of electrons $ = 2$
$m = \dfrac{F}{n} \times MW$
$m = \dfrac{{0.007}}{2} \times 56$
$m = 0.196g$
So the mass of gold and silver $ = $ mass of impurities $ - $ mass of iron
Mass of gold and silver $ = 0.249 - 0.196$
Mass of gold and silver $ = 0.053g$
D.We know the mass of copper and iron respectively. So now we will calculate the percentage of iron and copper present using the formula:
$\% = \dfrac{{mass}}{{total mass}} \times 100$
Given data: mass of copper $ = 22.011g$
Mass of iron $=0.196g$
Total mass $=22.260g$
i.$\% $of iron$ = \dfrac{{mass}}{{total mass}} \times 100$
Substituting the value we get,
$\% $ of iron$ = \dfrac{{0.196}}{{22.260}} \times 100$
$\% $ of iron$ = 0.88$
ii.$\% $ of copper$ = \dfrac{{mass}}{{total mass}} \times 100$
$\% $ of iron$ = \dfrac{{22.011}}{{22.260}} \times 100$
$\% $ of iron$ = 98.88$
So the percentage of iron and copper is $0.88 \% $ and $98.88\% $ respectively.

Note:
 Impurities such as $Fe,Cu,Zn,Ni$ get dissolved in the solution, whereas elements such as gold, silver, platinum get settled down as anode mud. Anode mud is the impurities that is collected at anode during the purification process of the metals.