Question

0.5 faraday of electricity was passed to deposit all the copper present in 500 ml of cuso4 solution. what was the molarity of this solution?

Answer 1

Answer:- The molarity of the copper sulfate solution is 0.5 M.

Solution:- When current is passed into a copper(II)sulfate solution then reduction of copper takes place. The reduction half equation is shown as:

$Cu^+^2(aq)+2e^-\rightarrow Cu(s)$

From this equation, 2 moles of electrons are used to deposit 1 mol of copper.

We know that, 1 mol of electron means one Faraday. So, 2 moles of electron is 2 faraday and it deposits 1 mol of copper.

Now, we can calculate the moles of copper got deposited by 0.5 faraday.

$0.5faraday*\frac{1 mol copper}{2 faraday}$

= 0.25 mol copper

These 0.25 moles of copper are present in 500 mL that is 0.5 L solution of copper sulfate.

So, the molarity of the solution = $\frac{0.25 mol}{0.5 L}$

= 0.5 M

Answer 2

Answer: 0.5M

Explanation:

Answer:- The molarity of the copper sulfate solution is 0.5 M.

Solution:- When current is passed into a copper(II)sulfate solution then reduction of copper takes place. The reduction half equation is shown as:

From this equation, 2 moles of electrons are used to deposit 1 mol of copper.

We know that, 1 mol of electron means one Faraday. So, 2 moles of electron is 2 faraday and it deposits 1 mol of copper.

Now, we can calculate the moles of copper got deposited by 0.5 faraday.

= 0.25 mol copper

These 0.25 moles of copper are present in 500 mL that is 0.5 L solution of copper sulfate.

So, the molarity of the solution =

= 0.5 M