The balanced half-reaction equation for the reduction of Pb2+ ions to lead metal is:
Pb2+ + 2e- → Pb
According to this equation, 2 electrons are required to reduce 1 Pb2+ ion to lead metal.
The molar mass of lead (Pb) is 207.2 g/mol, so the mass of 1 mole of lead is 207.2 g. Therefore, the mass of 0.299 g of lead is:
0.299 g / (207.2 g/mol) = 0.00144 mol
To deposit this amount of lead, we need to transfer 0.00144 mol of Pb2+ ions to lead metal. This requires the transfer of 2 x 0.00144 = 0.00288 moles of electrons.
The amount of electric charge required to transfer 0.00288 moles of electrons is:
Q = n x F = 0.00288 mol x 2 x 96,485.3329 C/mol = 557.4 C
where n is the number of moles of electrons transferred and F is the Faraday constant.
The amount of time required to deposit this amount of lead at a current of 0.627 A is given by the equation:
Q = I x t
where Q is the electric charge in coulombs, I is the current in amperes, and t is the time in seconds. Solving for t, we get:
t = Q / I = 557.4 C / 0.627 A = 888.9 seconds
Therefore, it would take approximately 888.9 seconds (or about 14.8 minutes) to deposit 0.299 g of lead metal from a solution containing Pb2+ ions at a current of 0.627 A.