Question

In the electrolysis of ${ CuCl }_{ 2 }$ solution, the mass of cathode increased by $6.4 g$. What occurred at Cu anode?
(A) 0.224 litre of ${ Cl }_{ 2 }$ was liberated.
(B) 1.12 litre of oxygen was liberated.
(C) 0.05 mole of ${ Cu }^{ 2+ }$ passed into solution.
(D) 0.1 mole of ${ Cu }^{ 2+ }$ passed into the solution.

Answer 1

For solving this question we need to compare the redox potentials for the reactions that can potentially take place at the cathode and the cathode to determine which species has a greater tendency to get reduced and oxidised respectively.

Complete step by step solution:
For solving this question we first need to understand Faraday’s first law of electrolysis which states that the amount of a substance that gets deposited on an electrode during electrolysis is directly proportional to the amount of charge passed through the electrolytic solution i.e.
${ mass\quad of\quad substance\quad deposited\quad \propto \quad Amount\quad of\quad charge }$
$\Rightarrow m=ZQ$
Where ‘m’ is the mass of the substance deposited, ‘Q’ is the amount of the charge passed through the solution and ‘Z’ is the proportionality constant; known as the electrochemical equivalent of the substance. ‘Z’ will be equal to the amount of the substance deposited when a charge of 1 coulomb is passed through the solution.
Let us examine the reaction given in the question:
We are given a solution of ${ CuCl }_{ 2 }$ (aqueous solution). At the cathode, the reduction will take place. There are two possible reduction processes that can take place. First is that the reduction of the ${ Cu }^{ 2+ }$ ions to $Cu$ and second is the reduction of the protons to give hydrogen gas (there will be some hydronium ions present in the aqueous solution due to self-ionization of water). The standard reduction potential for the reduction of ${ Cu }^{ 2+ }$ ions to $Cu$ is $0.337 V$ while the standard reduction potential for the reduction of ${ H }^{ + }$ to hydrogen gas is $0$. Hence ${ Cu }^{ 2+ }$ ions will get reduced at the cathode and get deposited at the cathode as $Cu$ metal.
Similarly, there are two possible oxidation processes that can take place at the anode. First is the oxidation of chloride ions to the chlorine gas which has a standard oxidation potential of $-1.36 V$. The second possible reaction is the oxidation of Cu metal of the anode to ${ Cu }^{ 2+ }$ ions which have a standard oxidation potential of $-0.34 V$. Since the oxidation potential for ${ Cu }^{ 2+ }$ ions to $Cu$ metal is less negative than the oxidation potential for chloride ions to chlorine gas, therefore, Cu metal of the anode will get oxidised to ${ Cu }^{ 2+ }$ ions which will go into the solution.
The two reactions are:
$\begin{matrix} At\quad anode: \\\ At\quad cathode: \end{matrix}\begin{matrix} Cu(s)\rightarrow { Cu }^{ 2+ }(aq)+2{ e }^{ - } \\\ { Cu }^{ 2+ }(aq)+2{ e }^{ - }\rightarrow Cu(s) \end{matrix}$
From the reactions, it is clear that the number of moles of Cu deposited on the cathode will be equal to the number of moles of copper that will get dissolved into the solution as ${ Cu }^{ 2+ }$ ions due to the oxidation reaction at the anode.
The molar mass of $Cu$ metal is $63.5 g$, therefore the number of moles of $Cu$ deposited on the cathode will be = $\cfrac { 6.4\quad g }{ 63.5\quad g/mol } =0.1\quad mol$

Hence the correct answer is (D) 0.1 mole of ${ Cu }^{ 2+ }$ passed into the solution.

Note: The redox potentials are actually the reduction potentials of the species and not the oxidation potentials so if the redox potential of a species is given, it is actually its reduction potential. Also if the redox potential of a species is more negative than the other, then the former has a less tendency to get reduced and the latter has a more tendency to get reduced.