How many minutes will be required to deposit 1.00 g of chromium metal from an aqueous?

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.

a

Explanation: