Question

A copper strip having 3cm length and 4cm breadth is coated with silver of 2mm thickness. Calculate the quantity of electricity (in coulomb) required for the above process. (Density of Ag = $10.5g{\text{ c}}{{\text{m}}^{ - {\text{3}}}}$)

Answer 1

The dimensions of silver are given which will give us volume of silver deposited, and density is also given, from above date we can find mass deposited of silver.
If we know mass deposited, then, we can find Quantity of current using Faraday’s First Law.
$m = Z.Q$
Z is electrochemical equivalent, Q is total charge or electricity flowing in the circuit.

Complete step by step answer:
We know from first law of electrolysis, the mass of silver deposited, can be given by formula:
$m = Z.Q$ --- equation 1
Q is the quantity of electricity which we have to find out.
To find mass (m), we know formula of density:
$density = \dfrac{{mass}}{{volume}}$
Density of silver = $10.5g{\text{ c}}{{\text{m}}^{ - {\text{3}}}}$ (given)
Length= 3cm; breadth = 4cm; thickness= 2mm= 0.2cm (given)
${\text{Volume = Length}} \times {\text{breadth}} \times {\text{thickness}}$
Substitute values to find volume.
$Volume = 3 \times 4 \times 0.2$
$\therefore Volume = 2.4c{m^3}$
Substitute, density, volume to get mass into formula:
$10.5 = \dfrac{{mass}}{{2.4}}$
So now to find mass, cross multiply.
$\therefore mass = 10.5 \times 2.4 = 25.2gm$
We need to find Z, electrochemical equivalent, using formula:
$Z = \dfrac{{{\text{Molar mass}}}}{{n \times F}}$
Molar mass = 108
F=96500 C
The reaction can be shown as below:
$A{g^ + } + {e^ - } \to Ag$
$\therefore n({\text{number of electrons) = 1}}$
Substitute all these values, to find Z:
$Z = \dfrac{{108}}{{1 \times 96500}}$
To substitute values of Z and mass into equation 1, we get:
$m = Z.Q$
$25.2 = \dfrac{{108}}{{96500}} \times Q$
Take all numerical values on one side, and find Q:
$Q = \dfrac{{25.2 \times 96500}}{{108}}$
$\therefore Q = 22516.67C$
Hence, the quantity of electricity (in coulomb) required for the above process is 22516.67 Coulomb.

Additional Information: Faraday’s first law of electrolysis says that the amount of substance produced at electrodes is directly proportional to the charge passing in the circuit.
$W\alpha Q$
Note:
Alternate Method: We also know the Atomic mass of Silver is 108, so we can find moles and use that formula which has relation of moles and charge required.
$n = \dfrac{Q}{{nf \times F}}$
Where, n= number of moles
$n = \dfrac{{mass}}{{{\text{Molar mass}}}}$
We know, mass= 25.2, Molar mass= 108, substitute values to get n,
$n = \dfrac{{25.2}}{{108}} = 0.233$
Q= Charge (We have to calculate)
nf= n-factor = 1 (since only 1 electron is involved)
F= Faraday constant=96500
Substitute these values in above equation, we get:
$0.233 = \dfrac{Q}{{1 \times 96500}}$
Taking Q on one side and numerical value on the other, we solve and simplify to get:
$Q = 0.233 \times 1 \times 96500$
$\therefore Q = 22516.67C$