Given:
1 g of chromium metal
CrO4 2- solution
6.0 amperes of current
Find:
Minutes required = ?
1 amp = 1 coulomb / second
6 amps = 6 coulombs / second...
Also:
Cr in CrO4(-2) is +6 so we must reduce it to 0, i.e. 6 electrons per atom
1g Cr x (1 mole Cr / 52.00g Cr) x (6 moles electrons / 1 mole Cr) x (6.022x10^23 electrons / mole electrons) x (1.602x10^-19 coulombs / 1 electron) x (1 sec / 6.00amps) x (1 min / 60x) = 30.9 min
Therefore, 30.9 minutes will be required to deposit 1.00 g of chromium metal from an aqueous CrO4 2- solution using a current 6.00 amperes.