Question

Find the mass of copper deposited during electrolysis of cuso4 solution using current of 2 ampere passed through the solution for 30 minutes

Answer 1

Solution :-

As per the given data ,

• Current (I) = 2A

• Molar mass of Cu (M) = 63.5 g

• Time = 30 min = 30 x 60 = 1800 s

• Faraday's constant (F) ≈ 96500

As per the formula ,

➝ W / E = Q / F

• W = mass of copper deposited
• E = Equivalent weight
• Q = Charge
• F = Faraday's constant

we know that ,

➝ E = M / z

here ,

• n = n - factor ( no of e- exchanged)

And ,

➝ Q = I x t

Now ,

➝ W x n / M = I x t  / F

On rearranging ,

➝ W = I x t x M / n x F

Substituting the given values in the above equation ,

➝ W = 2 x 1800 x 63.5 / 2 x 96500

➝ W = 1800 x 63.5 / 96500

➝ W = 114300 /96500

➝ W = 1.18 g

The mass of copper deposited during electrolysis is 1.18 g