Question

The mass of zinc produced by the electrolysis of zinc sulfate solution with a steady current of 0.015 A for 15 minutes is _____ \( \times \, 10^{-4} \) g.
(Atomic mass of zinc = 65.4 amu)

Answer 1

Correct Answer: 45.75

The reaction for the deposition of zinc is as follows:

Zn²⁺ + 2e^- → Zn

Using the formula for electrolysis:

W = \( \frac{Z \times i \times t}{F} \)

where

• Z = \( \frac{65.4}{2} \) (Equivalent weight of zinc),
• i = 0.015 A (current),
• t = 15 $\times$ 60 seconds,
• F = 96500 C/mol (Faraday constant).

Calculating the mass of zinc:

W = \( \frac{65.4}{2 \times 96500} \times 0.015 \times 15 \times 60 \)

W = 45.75 $\times$ 10^-4 g

Since the answer can be approximated, we also consider 46 $\times$ 10^-4 g.

So, the correct answer is: 45.75 or 46

Answer 2

Step 1: Use Faraday’s law of electrolysis.

According to Faraday’s first law: \[ m = \frac{E \, I \, t}{F} \] where \( m \) = mass of substance deposited (in g), \( E \) = equivalent weight of substance, \( I \) = current (A), \( t \) = time (s), \( F \) = Faraday’s constant \( = 96500\,\text{C/mol} \).

Step 2: Equivalent weight of zinc.

For Zn²⁺ + 2e⁻ → Zn, number of electrons \( n = 2 \). \[ E = \frac{\text{Atomic mass}}{n} = \frac{65.4}{2} = 32.7 \]

Step 3: Substitute given values.

\[ I = 0.015\,\text{A}, \quad t = 15\,\text{min} = 15 \times 60 = 900\,\text{s} \] \[ m = \frac{32.7 \times 0.015 \times 900}{96500} \]

Step 4: Simplify.

\[ m = \frac{32.7 \times 13.5}{96500} = \frac{441.45}{96500} = 0.00457\,\text{g} \] \[ m = 4.575 \times 10^{-3}\,\text{g} = 45.75 \times 10^{-4}\,\text{g} \]

Final Answer:

\[ \boxed{45.75 \times 10^{-4}\,\text{g}} \]