What current in ampere will deposit 0.27g of aluminium in 2 hours? [Al = 27, F = 96500C].

The amount of aluminum deposited on the electrode is directly proportional to the quantity of electricity passed through the cell. The formula for calculating the quantity of electricity is: Quantity of electricity = current x time We are given the mass of the aluminum and the time taken for it to be deposited, so we can use the formula for the amount of charge required to deposit that mass of aluminum: Quantity of electricity = (mass ÷ molar mass) x (1 Faraday ÷ 2 moles of electrons) = (0.27 ÷ 27) x (96500 ÷ 2) = 5052.5 C Now we can use the formula for current: Current = quantity of electricity ÷ time Plugging in the values we get: Current = 5052.5 C ÷ (2 x 3600 s) = 0.7 A Therefore, the current required to deposit 0.27 g of aluminum in 2 hours is 0.7 A. (0.4 A) is incorrect.