Question

Copper of thickness d is placed on the cathode of copper voltameter.if the total surface area of the cathode is 60cmsquare and steady current of 0.5ampere is maintained in the voltameter for 1 hour.calculate the value of d?

Answer 1

To calculate the value of d, we need to determine the amount of copper deposited on the cathode during the given time period.

First, let's find the amount of charge passing through the circuit in 1 hour:

Charge (Q) = Current (I) x Time (t)
Q = 0.5 A x 1 hour x 3600 seconds/hour = 1800 C

Since 1 Faraday (F) = 96500 C, we can find the amount of copper deposited on the cathode:

1 F of electricity deposits 1 mole of copper.

Amount of copper deposited = Charge / Faraday's constant
Amount of copper deposited = 1800 C / 96500 C/mol = 0.01867 mol

Now, let's calculate the mass of copper deposited:

Mass (m) = Number of moles x Atomic mass of copper
Mass = 0.01867 mol x 63.5 g/mol = 1.184 g

The surface area of the cathode is given as 60 cm^2, and since the thickness of the copper deposited is d, the volume of copper deposited is:

Volume = Area x Thickness
Volume = 60 cm^2 x d

Since the density of copper is 8.96 g/cm^3, we can calculate the mass of the deposited copper as:

Mass = Density x Volume
1.184 g = 8.96 g/cm^3 x 60 cm^2 x d

Solving for d:

1.184 g = 537.6 g/cm x d
d = 1.184 g / 537.6 g/cm
d ≈ 0.0022 cm

Therefore, the value of d is approximately 0.0022 cm.