Question

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol^–1 , 1 F = 96487 C)

• A. 3.15 g
• B. 0.315 g
• C. 31.5 g
• D. 0.0315 g

Answer 1

The Correct Option is B

To determine the mass of copper deposited by passing a current of 9.6487 A for 100 seconds, Faraday's laws of electrolysis are applied. These laws establish a direct proportionality between the mass of substance deposited and the total electric charge passed through the solution. The relationship is expressed as:

n = Q / 96500

Here, n represents the number of moles of electrons, Q is the total electric charge in coulombs, and 96500 is the Faraday constant (approximately 96487 C).

Step 1: Calculate the total charge (Q).

The total charge Q is computed using the formula:

Q = I × t

where I is the current in amperes and t is the time in seconds.

Given I = 9.6487 A and t = 100 s:

Q = 9.6487 × 100 = 964.87 C

Step 2: Calculate the moles of copper deposited.

The reduction of Cu²⁺ to Cu necessitates 2 moles of electrons for every mole of copper deposited. Thus:

n (Cu) = 964.87 / (2 × 96487)

n (Cu) = 0.005

Step 3: Calculate the mass of copper deposited.

The mass (<i>m</i>) is calculated using the molar mass (<i>M</i>) with the formula:

m = n × M

For Cu, the molar mass M is 63 g/mol.

m = 0.005 × 63 = 0.315 g

Consequently, the mass of copper deposited is 0.315 grams.

Conclusion:

The calculated mass of deposited copper is 0.315 g.