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

Answer:

The charge on the cation = 3

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 =

$\frac{1}{96500}$

And 0.15 x  12600 C of electron contain:

$\frac{1}{96500}\times 0.15\times 12600$ 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:

$Moles =\frac{mass}{Molar\ mass}$

$Moles =\frac{0.176}{27}$

$Moles =0.00652$moles

Now , the moles transferred (Magnitude of charge):

$Charge\ transferred=\frac{Moles\ in\ electrons}{Moles}$

$Charge =\frac{0.0195}{0.00652}$

Charge transferred = 2.99(Nearly equal to 3 )

or

Charge transferred = 3