What current will deposit 3.25g of zinc in 2 hrs? [Zn=65, F=96500Cmol^-1]

What current will deposit 3.25g of zinc in 2 hrs? [Zn=65, F=96500Cmol^-1]

A. 3.25A

B. 2.00A

C. 1.34A

D. 0.67A

Let’s apply Faraday’s law of electrolysis step by step.

Step 1: Write the electrode reaction

Zn^2+ + 2e^- 一> Zn

So, 2 moles of electrons are needed to deposit 1 mole of Zn.

Step 2: Calculate moles of zinc deposited

Moles of Zn = 3.25/65 = 0.05 mol

Step 3: Calculate total charge required

Q = n * z * F
Where:

• (n = 0.05) mol

• (z = 2)

• (F = 96500 C mol^-1

Q = 0.05 * 2 * 96500 = 9650 C

Step 4: Convert time to seconds

2 hours = 2 * 3600 = 7200 s

Step 5: Calculate the current

I = Q/t = 9650/7200 = 1.34  A

✅ Correct answer: C. 1.34 A

Exam tip:
I = (m *z * F)/(M * t)
PQN: 33