Question

A certain current liberates 0.5 g of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?
(A) 12.7g
(B) 15.9 g
(C) 31.8g
(D) 63.5 g

Answer 1

Hint: In order to solve the question, we need to first know the moles of copper produced. Then knowing the atomic weight of Cu, will give us the grams of Cu produced.

Complete step by step solution :

Let us first write the equation which is mentioned in the question:
$\text{2}{{\text{H}}^{\text{+}}}\text{+ 2}{{\text{e}}^{-}}\text{ }\xrightarrow{{}}\text{ }{{\text{H}}_{\text{2}}}$
$\text{Mole ratio of hydrogen = }\dfrac{\text{moles of hydrogen produced in the half reaction}}{\text{moles of electrons required in the half reaction}}$
- Mole ratio of hydrogen = $\dfrac{1}{2}$
Now let us write the equation of Cu:
$\text{C}{{\text{u}}_{\text{2}}}^{\text{+}}\text{ }\\!\\!~\\!\\!\text{ + 2}{{\text{e}}^{-}}\text{ }\xrightarrow{{}}\text{ Cu}$
$\text{Mole ratio of Cu = }\dfrac{\text{moles of Cu produced in the half reaction}}{\text{moles of electrons required in the half reaction}}$
- Mole ratio of hydrogen = $\dfrac{1}{2}$
$\dfrac{\dfrac{0.50}{2}}{\text{moles of copper produced}}\text{ }=\text{ }\dfrac{\dfrac{1}{2}\text{ }}{\dfrac{1}{2}}$
- Moles of copper produced $\text{= 0}\text{.25}$
We know the atomic weight of Cu is 63.5
Now we have to find the mass of copper produced.
We know that mass of copper will be moles of copper produced multiplied by atomic weight of copper.
So, Mass of copper produced $\text{0}\text{.25 }\\!\\!\times\\!\\!\text{ 63}\text{.5 = 15}\text{.9 g}$
Therefore 15.9 grams of copper is liberated by the same current flowing for the same time in a copper sulphate solution.
So the correct answer is Option B.

Note: In order to find the atomic weight of any atom it can be found by multiplying the abundance of an isotope of an element by the atomic mass of the element and then adding the results together.There are two isotopes of copper available $^{\text{63}}\text{Cu}$ and $^{\text{65}}\text{Cu}$.