Question

What mass of gold is produced when 17.6^A of current are passed through a gold solution for 37.0 min ?

Answer 1

Answer:

There is 26.58 grams of gold formed

Explanation:

Step 1: Data given

17.6 A of current are passed through a gold solution for 37.0 min

Molar mass of Au = 196.967 g/mol

Step 2: The equation

Au^3+ + 3e- → Au

Step 3: Calculate coulombs

17.6 Coulomb/s * 37.0 min * 60 sec/min = 39072 Coulombs

1 Faraday = 96500 Coulombs

Step 4: Calculate faraday

39072 Coulombs / 96500 Coulombs / Faraday = 0.40489 Faraday

Step 5: Calculate mass of gold formed

For every 3 Faraday of electricity used up , 1 mole Au is formed

0.40489 Faraday * 1 mole Au/ 3 Faraday = 0.13496 mole Au

 

196.967 g/mol * 0.13496 mol = 26.58 g Au

There is 26.58 grams of gold formed

Answer 2

The mass of gold that is produced is 26.59 g

Using the formula

$m = \frac{Atomic\ mass}{nF}\times It$

Where m is the mass

n is the number of equivalents

F is the Faraday constant ( F = 96485 C)

I is the current

and t is the time

From the given information

I = 17.6 A

t = 37.0 min = 37.0 × 60

t = 2220 secs

For gold

Atomic mass = 196.97 g/mol

and n = 3

Putting these parameters into the formula, we get

$m = \frac{196.97}{3 \times 96485} \times 17.6 \times 2220$

$m = \frac{7696011.84}{289455}$

m = 26.59 g

Hence, the mass of gold that is produced is 26.59 g

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