Electrolysis Calculation: Copper Deposited Mass

This problem involves calculating the mass of copper deposited at the cathode during the electrolysis of a copper sulphate solution using Faraday's laws of electrolysis.

Step 1: Convert Time to Seconds

The time given is $10 \text{ minutes}$. To use it in the charge calculation, convert it to seconds:

Time ($t$) = $10 \text{ minutes} \times 60 \frac{\text{seconds}}{\text{minute}} = 600 \text{ seconds}$

Step 2: Calculate Total Charge Passed

The charge ($Q$) passed is the product of current ($I$) and time ($t$):

$Q = I \times t$

Given $I = 1.5 \text{ amperes}$ and $t = 600 \text{ seconds}$:

$Q = 1.5 \text{ A} \times 600 \text{ s} = 900 \text{ Coulombs (C)}$

Step 3: Determine Moles of Electrons and Copper

The deposition of copper from copper sulphate solution ($CuSO_4$) involves the reduction of copper ions ($Cu^{2+}$) at the cathode:

$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

This reaction shows that 2 moles of electrons ($e^-$) are required to deposit 1 mole of copper ($Cu$).

First, find the number of moles of electrons passed using Faraday's constant ($F = 96487 \text{ C mol}^{-1}$):

Moles of electrons = $\frac{\text{Total Charge (Q)}}{\text{Faraday's Constant (F)}} = \frac{900 \text{ C}}{96487 \text{ C mol}^{-1}} \approx 0.0093276 \text{ mol}$

Now, calculate the moles of copper deposited:

Moles of $Cu$ = $\frac{\text{Moles of electrons}}{2} = \frac{0.0093276 \text{ mol}}{2} \approx 0.0046638 \text{ mol}$

Step 4: Calculate Mass of Copper Deposited

The mass of copper deposited is the product of the moles of copper and its molar mass:

Mass of $Cu$ = Moles of $Cu$ $\times$ Molar Mass of $Cu$

Given Molar Mass of $Cu = 63 \text{ g mol}^{-1}$:

Mass of $Cu = 0.0046638 \text{ mol} \times 63 \text{ g mol}^{-1} \approx 0.2938194 \text{ g}$

Rounding to four decimal places, the mass deposited is approximately $0.2938 \text{ g}$.