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# Calculate the mass of copper deposited when a current of 0.5 ampere was passed t...

### Question

Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]

A) 0.300g

B) 0.250g

C) 0.2242g

D) 0.448g

### Explanation:

M = $$\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}$$

= $$\frac{MmIT}{96500n}$$

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = $$\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}$$

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = $$\frac{64 \times 1350}{96500 \times 2}$$

Mass = 0.448g

## Discussion (2)

• The equation for the electrolytic reaction is;
$$CuCl_2\rightarrow Cu^{2+} +2Cl^{-}$$
Mass =[M]
Molecular mass [Mm]=$$64gmol^{-1}$$
Current [I]= 0.5A
Time [T] = 45mins= 45 × 60= 2700sec
charge [n]=2
$$M=\frac{MmIt}{96,500n}$$
$$M= \frac{64×0.5×2700}{96500×2}$$
$$M= \frac{86,400}{193,000}$$
M=0.4477g
M≈ 0.448g
Option [D] of s very much correct.

• M = $$\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}$$

= $$\frac{MmIT}{96500n}$$

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = $$\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}$$

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = $$\frac{64 \times 1350}{96500 \times 2}$$

Mass = 0.448g