Gold (atomic mass = 197 g/mol) is plated from a solution of chlorauric acid, HAuCl₄; it deposits on the cathode. Calculate the time it takes to deposit 0.65 g of gold, passing a current of 0.14 amperes. (1 faraday = 96,485 coulombs)

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of charge required to deposit 0.65 g of gold:

1 mole of gold = 197 g
0.65 g of gold = 0.65/197 moles of gold

1 mole of gold requires 3 electrons to be deposited
Therefore, 0.65/197 moles of gold requires 0.65/197 x 3 electrons

The charge required = 0.000629 C

Now, we can use Faraday's Laws of Electrolysis to calculate the time required:

Q = It, where Q is the charge, I is the current, and t is the time.

1 Faraday = 96,485 C
Therefore, the time required, t = (Q/I) x (1/96485)

Substituting the values, we get:

t = (0.000629/0.14) x (1/96485) = 0.00184 hours or 1.9 hours (approx.)

Therefore, the answer is C) 1.9 h.

Gold (Atomic Mass = 197 G/mol) Is Plated from a Solution