Question

A solution of copper sulphate electrolysed for 20 minutes with a current of 1.5 Ampere. Calculate the mass of copper deposited at the cathode. (F = 96500C)

Answer 1

A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes.

Answer 2

The mass of copper deposited at the cathode is 0.59 g

Explanation:

1 electron carry charge=$1.6\times 10^{-19}C$

1 mole of electrons contain=  $6.023\times 10^{23}$ electrons

Thus  1 mole of electrons carry charge= $1.6\times 10^{-19}\times 6.023\times 10^{23}=96500C$

If n electrons are involved in the reaction, the passage of n Faradays i.e  of electricity will liberate 1 mole or molecular mass of the substance.

$I=\frac{q}{t}$

I = current passed = 1.5 A

q = total charge = ?

t = time required = 20 min = 1200 sec    (1 min= 60 sec)

Putting values in above equation, we get:

$q=1.5\times 1200=1800C$

At cathode :

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ deposits = 63.5 g of copper

1800 C deposits =$\frac{63.5}{193000}\times 1800=0.59g$

Learn more about Faraday law

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