If 10.8 g of silver is deposited in a silver coulometer volume of oxygen liberated is?
0.56 dm3
5.60 dm3
11.20 dm3
22.40 dm3
Correct answer is A
Ag+ + e-
H = 108 g Ag
= 0.11 10.8 Ag
4OH → O2 + 2H2O + 4e
4F ↔ 1mole O2
4F = 22.4 dm3 χ 0.1 = 0.56 dm3