Question

Calculate the time to deposit 1.27 g of Copper at Cathode When a current of 2A. Was passed through the sol^n of CuSO₄. (Molar Mass of Cᵤ = 63.5 g mol, F = 96500 Cmol⁻¹)

Answer 1

1930 seconds

Answer 2

To calculate the time required to deposit copper, we use Faraday's laws of electrolysis.

1. Determine the half-reaction for copper deposition:

At the cathode, copper(II) ions are reduced to metallic copper. $\text{Cu}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Cu}(s)$

From this reaction, 2 moles of electrons are required to deposit 1 mole of copper. Therefore, the valency (n-factor) for copper is 2.

2. Calculate the equivalent mass (E) of Copper:

The equivalent mass is given by: $E = \frac{\text{Molar Mass}}{\text{Valency}}$

Given Molar Mass of Cu = 63.5 g/mol $E = \frac{63.5 \text{ g/mol}}{2 \text{ eq/mol}} = 31.75 \text{ g/eq}$

3. Apply Faraday's First Law of Electrolysis:

The mass of a substance deposited ( m ) is related to the current ( I ), time ( t ), equivalent mass ( E ), and Faraday's constant ( F ) by the formula: $m = \frac{E \times I \times t}{F}$

We need to find the time ( t ), so we rearrange the formula: $t = \frac{m \times F}{E \times I}$

Given values:

Mass of Copper ( m ) = 1.27 g Current ( I ) = 2 A Faraday's constant ( F ) = 96500 C/mol (or C/eq)

Substitute the values into the formula: $t = \frac{1.27 \text{ g} \times 96500 \text{ C/mol}}{31.75 \text{ g/eq} \times 2 \text{ A}}$ $t = \frac{122555}{63.5}$ $t = 1930 \text{ seconds}$

The time required to deposit 1.27 g of Copper is 1930 seconds.