Question

A certain current liberated 0.504 g of hydrogen in 2 hours. The weight of Cu deposited by the same current flowing for the same time in copper sulphate solution in gram is:

Answer 1

To answer this question, we should first know about Faraday’s law of electrolysis. It is important that we should focus on Faraday’s second law.

Complete step by step answer:

Let us first start the answer of this question by focusing on Faraday’s law of electrolysis. We should know that Faraday's laws are primary laws of electrolysis.

Faraday’s first law states that during electrolysis, the amount of chemical reaction which occurs at any electrode under the influence of electrical energy is proportional to the quantity of electricity passed through the electrolyte.

Now, we will discuss Faraday's second law. Second law states that when quantity of electricity passes through the electrolytic solution, a number of different substances liberated are proportional to their chemical equivalent weights.

Let us take one example:

$\text{N}{{\text{a}}^{+}}+{{\text{e}}^{-}}\to \text{ Na}$

In the above reaction, we can say that one mole of the electron is required for the reduction of one mole of sodium ions. We know that charge on one electron is equal to $1.6021 \times 10^{–19}$C. Therefore, the charge on one mole of electrons is equal to:

${{\text{N}}_{\text{A }}}\times \text{ 1}.\text{6}\times \text{ 1}{{0}^{\text{19}}}\text{C }$

$\text{ 6}.0\text{2 }\times \text{ 1}{{0}^{\text{23}}}\text{mo}{{\text{l}}^{\text{1}}}\text{ }\times \text{ 1}.\text{6 }\times$ $\text{ 1}{{0}^{\text{19}}}\text{C }=\text{ 96487 C mo}{{\text{l}}^{\text{1}}}$

We should know that, quantity of electricity is defined as one Faraday and is denoted by F.

Now, we will calculate our answer:

We will use the Faraday’s second law:

Weight deposited Equivalent weight

We can write this as:

$\dfrac{\text{weight of Cu}}{\text{weight of }{{H}_{2}}}=\dfrac{\text{Equivalent weight of Copper}}{\text{Equivalent weight Hydrogen}}$

Equivalent weight is defined as the ratio of the atomic mass of metal and the number of electrons required for reducing the cation.

$\dfrac{\text{weight of Cu}}{1}=\dfrac{\dfrac{63.5}{2}}{1}$

After this we can say that, weight of copper is 31.8 gram.

Therefore weight of Cu deposited is 31.8 gram.

Note:
We should know that electrolysis is a method of removing iron oxide by passing a small electrical charge through the rusty metal from a battery or battery charger to start ion exchange while the device is submerged in an electrolyte solution.