Question

How many grams of $Cu$ and what volume of ${O_2}$ gas will be obtained at STP by passing 20 ampere current for $96.5$ minutes from an aqueous solution of $CuS{O_4}$? The efficiency of the electrolytic cell is $90\% (Cu = 63.5)$.

Answer 1

Hint: Electrolysis (Electro-electricity, lysis-breaking) is the process of breaking chemical compounds called Electrolytes $(NaCl,N{a_2}S{O_4}etc.)$ in the presence of electricity in the form of electric current in the Electrolytic cell. Oxidation occurs at Anode and Reduction occurs at Cathode thus Cations get attracted to Cathode and Anions get attracted to the Anode.

Complete answer:
Faraday’s Law of Electrolysis-
The mass of substance deposited or mass of gas liberated at a particular electrode is proportional to the amount of charge passing through the solution.
$\begin{gathered}
  m \propto Q \\
  m = ZQ - - - - (1) \\
\end{gathered} $
Where $m$ is the mass, $Q$ is the charge and $Z$ is a constant called Electrochemical Equivalent.
We can replace the value of $Q$ in equation (1) by $it$ where $i$ is charged and $t$ is the time.
$m = Zit - - - - (2)$
Electrochemical Equivalent $Z$ is given as Equivalent mass/Faraday’s Constant or$\dfrac{E}{F}$.
Here 1 Faraday is charged on 1 mole of electron. 1 Faraday=96500 C/mole.
Thus, $Z = \dfrac{E}{F} - - - - (3)$
Equivalent Mass=Molar mass/n factor where n factor is basicity, acidity or overall positive or negative charge in acid, base and salt respectively.
Put equation (3) in equation (2), $m = \dfrac{E}{F}it - - - - (4)$
In the question we are given the Electrolysis of Aqueous Copper Sulphate solution-
Current $i = 20ampere$
Time $t = 96.5\min = 96.5 \times 60\sec = 5790\sec $
The reactions occurring at cathode and Anode are as follows-
Cathode-$C{u^{ + 2}} + 2{e^ - } \to Cu - - - - (5)$
Anode-$4O{H^ - } \to 2{H_2}O + {O_2} + 4{e^ - } - - - - (6)$
Molar mass$ = 63.5$ and copper is deposited in the form of $C{u^{ + 2}}$ so the value of n factor will be equal to 2.
So Equivalent mass$ = \dfrac{{63.5g/mole}}{2} = 31.75g/mole$
The mass of copper obtained will be $m = \dfrac{{31.75(g/mole)}}{{96500(C/mole)}} \times 20(A) \times 5790(\sec ) = 38.1g$
The charge$(Q)$ in case of Oxygen is calculated as $ = 20(A) \times 5790(\sec ) = 115800C$
4 moles of electrons gives 1 mole of oxygen gas. (As seen by equation (6))
If 4 mole produces $22.4L$ of Oxygen (At STP 1 mole of any gas occupies $22.4L$ ) or $4 \times 96500C$ produces 22.4 litres of oxygen,
Then 115800 C produces $ = \dfrac{{115800C \times 22.4L}}{{4mole \times 96500C/mole}} = 6.72L$
However in the question we are specified that efficiency of electrolytic cells is $90\% $,
So Mass of Copper obtained$ = 38.1g \times \dfrac{{90}}{{100}} = 34.3g$
Volume of Oxygen obtained$ = 6.72L \times \dfrac{{90}}{{100}} = 6.05L$

Note:
 Electrolysis of Aqueous copper sulphate results in deposition of Copper at the Cathode and release of Oxygen gas takes place at the Anode because here water is used as a solvent. In the numerical problems it is necessary to write the half-cell reactions first and also it is important to convert time in minutes to seconds in order to avoid any error.