Question

a metal of atomic mass 27g is deposited by electrolysis. If 0.176 of the metal is deposited on a cathode when 0.15A flows for 3.5hours. what is the magnitude of the charge on the cations​

Answer 1

The charge on the cation = 3

Step-by-step explanation:

According to Faraday's law of electrolysis , the mass of the substance deposited or liberated at any electrode is directly proportional to quantity of electricity passed.

Q= zIt

The charge can be calculated using :

Q = It

Here I = current in Ampere(A)

t = time in seconds(s)

Q = Charge in coulomb(C)

1 hours = 60 min and 1 minute = 60 second

1 hours = 60 x 60 second = 3600 sec

1 hrs = 3600 s

3.5 hrs = 3600 x 3.5 sec

= 12600 sec

I =0.15 A

t = 12600 sec

Q = It

Q = 0.15 x 12600 C

Now, the charge carried by 1 mole of electricity = 96500 C

Or,

96500 C is present in = 1 mole of electricity

1 C of charge is present in =

`(1)/(96500)`

And 0.15 x 12600 C of electron contain:

`(1)/(96500)* 0.15* 12600` moles

`=0.0195`moles of electrons

The Molar mass of the substance = 27 gram

The mass deposited = 0.176 gram

The moles deposited by the substance can be calculated by the following formula:

`Moles =(mass)/(Molar\ mass)`

`Moles =(0.176)/(27)`

`Moles =0.00652`moles

Now , the moles transferred (Magnitude of charge):

`Charge\ transferred=(Moles\ in\ electrons)/(Moles)`

`Charge =(0.0195)/(0.00652)`

Charge transferred = 2.99(Nearly equal to 3 )

or

Charge transferred = 3