Question

A certain current liberates $0.5g$ of hydrogen in $2\text{ hours}$. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?

Answer 1

Hint It is based on the Faraday’s second law of electrolysis and in this first, we have to find the equivalent weight of the hydrogen and the copper and then, by using them in the formula: $\dfrac{\text{ weight of }{{\text{H}}_{2}}}{\text{ equivalent weight of }{{\text{H}}_{2}}}=\dfrac{\text{weight of Cu}}{\text{ equivalent weight of Cu}}$, we can easily find the weight of the copper in the solution. Now solve it.

Complete step by step answer:
First of all, let’s discuss Faraday's law of electrolysis. It states that when some quantity of electricity is passed through different electrolytic solutions connected in series, the weights of the substances produced at the electrodes are directly proportional to their chemical equivalent weights.
Now considering the numerical;
$\dfrac{\text{ weight of }{{\text{H}}_{2}}}{\text{ equivalent weight of }{{\text{H}}_{2}}}=\dfrac{\text{weight of Cu}}{\text{ equivalent weight of Cu}}$ -------------(1)
As we know that weight of hydrogen is $0.5g$(given)
And suppose that the weight of copper is $x\text{ g}$.
We can find the equivalent weight of the hydrogen by applying the formula as;
$equivalent\text{ }weight=\dfrac{atomic\text{ mass}}{2}$
We know that the atomic mass of hydrogen is 2.
Then, the equivalent weight of hydrogen is;
$equivalent\text{ }weight=\dfrac{2}{2}=1$
Similarly, we can find the equivalent weight of the copper.
the atomic mass of copper is $63.5$ .
Then, the equivalent weight of copper is;
$equivalent\text{ }weight=\dfrac{63.5}{2}=31.75$
Now put these all, values in equation (1), we get;
$\begin{align}
& \dfrac{0.5}{\text{ 1}}=\dfrac{x}{31.75} \\
& x=0.5\times 31.75 \\
& x=15.87 \\
& x\approx 15.9\text{ g} \\
\end{align}$
Thus, when the current liberates $0.5g$ of hydrogen in $2\text{ hours}$, then $15.9\text{ g}$ of copper is liberated by the same current flowing for the same time in a copper sulphate solution.

Note: In terms of moles of electrons, Faraday’s second law of electrolysis may be defined as; the moles of substances deposited or liberated is directly proportional to the number of moles of electrons exchanged during the oxidation-reduction reactions that occur.