0.414 g of lead was deposited at the cathode when a current of 2.5 A was passed. (Ar Pb = 207; 1 Faraday = 96 500 coulombs)

1. aCalculate the number of moles in 0.414 g of lead. [1]
2. bCalculate the total charge in Faradays required to liberate 0.414 g of lead at the cathode. [2]
3. cCalculate the amount of charge in coulombs. [1]
4. dCalculate the time taken, in seconds, to liberate the 0.414 g of lead if the current is 2.5 A. [2]

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