Through a ZnSO₄ solution, 0.015 A current was passed for 15 minutes. What is the mass of Zn deposited? (in mg) (Atomic weight of Zn = 65.4)

The correct answer is 4.58 mg (or 5 mg if rounded to the nearest integer).

Step 1: Understand the Given Parameters

Current ($I$) = 0.015 A
Time ($t$) = 15 minutes = 15 × 60 seconds = 900 s
Molar mass of Zinc (Zn) = 65.4 g/mol
Faraday's constant ($F$) = 96500 C/mol

Step 2: Calculate Total Electric Charge Passed

The total electric charge ($Q$) passed through the solution is given by Faraday's first law of electrolysis:

$Q=I\times t$

$Q=0.015\times 900=13.5\text{ C}$

Step 3: Determine Moles of Electrons Transferred

The number of moles of electrons passed through the circuit is calculated using Faraday's constant:

$\text{Moles of electrons}=\frac{Q}{F}$

$\text{Moles of electrons}=\frac{13.5}{96500}\approx 0.0001399\text{ mol}$

Step 4: Determine Moles of Zinc Deposited

In a ZnSO₄ solution, zinc exists as Zn²⁺ ions. The reduction reaction at the cathode is:

Zn²⁺ + 2e^- → Zn

This stoichiometry shows that 2 moles of electrons are required to deposit 1 mole of Zn atoms. Thus:

$\text{Moles of Zn deposited}=\frac{1}{2}\times \text{Moles of electrons}$

$\text{Moles of Zn deposited}=\frac{1}{2}\times 0.0001399=0.00006995\text{ mol}$

Step 5: Calculate Mass of Zinc Deposited

Now, multiply the number of moles of Zn by its atomic weight to find the mass in grams:

$\text{Mass of Zn}=0.00006995\times 65.4=0.004575\text{ g}$

Converting grams to milligrams (1 g = 1000 mg):

$\text{Mass of Zn}=0.004575\times 1000=4.58\text{ mg}$

Thus, the mass of Zn deposited is 4.58 mg (which rounds to 5 mg).