If 10.8 g of silver is deposited in a silver coulometer volume of oxygen liberated is?
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Correct Answer: Option A
Explanation:
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
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