Question

If a piece of jewelry were electroplated with gold for 25 seconds at 1.5 A, how many grams of gold would be plated? (Note: 1 A = 6.241 1018 e–/s. The reaction for the reduction of gold ions to gold metal is Au+ + 1e– Au(s).)

Answer 1

7.7x10^-2 grams
   Since the reaction indicates that 1 electron will result in the plating of 1 atom of gold, let's figure out how many electrons we get with 25 seconds of current at 1.5 amperes. An ampere is defined as 1 coulomb per second, so we have: 1.5 * 25 = 37.5 coulombs of electrons.
   A coulomb of electrons is about 6.241Ă—10^18 electrons. So the number of electrons we have is: 6.241Ă—10^18 * 37.5 = 2.340375x10^20
   So we will have plated 2.340375x10^20 atoms of gold. Let's figure out how many moles that is by dividing by Avogadro's number: 2.340375x10^20 / 6.0221409x10^23 = 3.886284x10^-4
   Finally, multiply the number of moles plated by the atomic weight of gold to get the number of grams. So 3.886284x10^-4 * 196.96657 = 7.6546804x10^-2
   So, rounding to 2 significant figures gives 7.7x10^-2 grams, or 77 milligrams of gold.