Question

A current of 2.68 ampere is passed for one hour through an aqueous solution of copper
sulphate using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At.
Mass of copper $=$ 63.5)

Answer 1

This question is based on the concept of electrolysis of copper sulphate solution using copper
electrode. In this a reaction will occur at cathode and at anode. When an electric current is passed
electrons will get attracted to the cathode also known as negative terminal of the battery and ions
will be deposited on the cathode. The formula used to calculate the number of faradays is
=$\dfrac{{It}}{{96500}}$

Complete step by step answer:
Now, first we will calculate the number of faradays by using the mentioned formula.
Here, $\dfrac{{It}}{{96500}}$ I represents the current passed, and t represents the time.
Substituting the values, we get
Number of faradays $= \dfrac{{2.68 \times 1 \times 3600}}{{96500}}$
Number of faradays $= 0.1$
The reaction of copper sulphate solution using copper electrode at cathode and anode can
be written as:
At cathode: $C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)$
At anode: $Cu(s) \to C{u^{2 + }}(aq) + 2{e^ - }$
We can say that the mass of copper will increase at cathode as the copper is getting
deposited whereas at anode mass will decrease as the copper is getting dissolved.
Now, let us calculate the mass deposited at cathode. The reaction is
$C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)$
Here, $\text{2F = 1 mole of copper}$
Then, $\text{0.1F = 0.05 mole of Copper}$
$\Rightarrow 0.1F = 0.05 \times 63.5g Copper$ (Atomic mass of copper is 63.5 g)
$\therefore 0.1F = 3.175g$
Now, we can conclude that at cathode 3.175 g of copper is deposited
So, now we will calculate the mass getting dissolved at anode. The reaction is
$Cu(s) \to C{u^{2 + }}(aq) + 2{e^ - }$
Here, we have $\text{2F = 1 mole of copper}$
Then, $\text{0.1F = 0.05 mole of Copper}$
$\Rightarrow 0.1F = 0.05 \times 63.5g Copper$ (Atomic mass of copper is 63.5 g)
$\therefore 0.1F = 3.175g$
So, the amount used of copper is 3.175 g.
In the last, we can conclude that mass deposited at the cathode is equal to the mass
dissolved at the anode. The value of mass is 3.175 g.
Thus the correct option is D.

Note:
There is an alternate method to find the mass of copper at the cathode and anode. We have
already mentioned the reactions occurring at cathode and anode.
Therefore, Charge passed through the electrode $= 2.68 \times 60 \times 60$, (the formula used
$[Q = It]$)
Copper that is deposited or dissolved $= \dfrac{{63.52}}{{2 \times 96500}}2.68 \times 60 \times 60$
Mass of copper deposited or dissolved is 3.175 g.
This method can also be used.