Answer:
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
