Question

How many electrons would be required to deposit $6.35gm$ of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate?
[Atomic mass of copper$ = 63.5u$,${N_A} = $avogadro’s constant]
A: $\dfrac{{{N_A}}}{{20}}$
B: $\dfrac{{{N_A}}}{{10}}$
C: $\dfrac{{{N_A}}}{5}$
D: $\dfrac{{{N_A}}}{2}$

Answer 1

Hint: In one mole of a substance there is an Avogadro number of particles. Mass of one unit of substance is calculated by dividing molecular mass with Avogadro's number. In the process of electrolysis there are two electrodes, cathode and anode.

Complete step by step answer:
In this question we have to find the electrons that would be required for the production of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate. Formula of copper sulphate is $CuS{O_4}$. Charge on sulphate ion $\left( {SO_4^{2 - }} \right)$ is $ - 2$. As a whole, the compound is neutral. Therefore charge on copper ions is $ + 2$. In the process of electrolysis there are two electrodes, cathode and anode. Anode is that electrode where oxidation occurs and cathode is that electrode where reduction occurs. Oxidation is defined as the loss of electrons and reduction is defined as the gain of electrons. In this question reaction is taking place at cathode. This means reduction will occur. Copper has $ + 2$ charge in its compound. This means copper ions can gain two electrons to form copper. Reaction that will take place is:
$C{u^{2 + }} + 2{e^ - }\xrightarrow{{}}Cu$
From this reaction we can conclude that two moles of electrons are required to form one mole of copper. One mole of electrons will contain Avogadro number of particles. This means to produce one mole of copper two times Avogadro number of particles are required. Mass of one mole of copper is given that is $63.5g$. So, this can be written as,
$63.5g{\text{ of copper}} = \left( {2 \times {N_A}} \right){\text{electrons}}$
$1g{\text{ of copper}} = \left( {\dfrac{{2 \times {N_A}}}{{63.5}}} \right){\text{electrons}}$
In this question we have to find the electrons required for production of $6.35gm$ of copper. So,
$6.35g{\text{ of copper}} = \left( {\dfrac{{2 \times {N_A} \times 6.35}}{{63.5}}} \right){\text{electrons}}$
Solving this we get,
$6.35g{\text{ of copper}} = \left( {\dfrac{{{N_A}}}{5}} \right){\text{electrons}}$
So, the correct answer is option C.

Note:
In this question reaction was at cathode. Cathode is the electrode where reduction occurs. Reduction is defined as the gain of electrons. if the reaction would be at cathode then there would be oxidation of given substance.