Calculate the amount of gold deposited when a current of 5A is passed through a solution of gold salt for 2hrs 15mins. If the same current is used, find the time taken for 6g of gold to be deposited. (Au=197 1faraday=96500c)

a. 82.68 g b. 9.8 min

Explanation:

a. The amount of gold deposited by 5 A current in 2 hrs 15 mins

Since charge Q = It where I = current = 5 A and time = 2 hrs 15 mins = 2 × 60 min + 15 min = 120 + 15 min = 135 min = 135 × 60 s = 8100 s

Q = 5 A × 8100 s

= 40500 C

Also, Q = nF where n = number of moles of gold deposited and F = Faraday's constant = 96500 C

n = Q/F = 40500 C/96500 C = 0.4195 moles ≅ 0.42 mole

Now n = m/M where m = mass of gold and M = molar mass of gold = 197

m = nM

= 0.42 × 197 g

= 82.68 g

b. The time taken for 6g of gold to be deposited.

We first find the number of moles of gold in 6g of gold

Since n = m/M and m = 6 g

n = 6/197 = 0.0305 mole

Q = It = nF

t = nF/I

= 0.0305 mol × 96500 C/5 A

= 2939.09 mol C

= 587.82 s

Changing t to minutes

587.82/60 s = 9.8 min

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