Question 19.10: Stoichiometry of Electrolysis Gold can be plated out of a so...
Stoichiometry of Electrolysis
Gold can be plated out of a solution containing Au^{3+} according to the half-reaction:
Au^{3+} (aq) +3 e^- \longrightarrow Au(s)
What mass of gold (in grams) is plated by a 25-minute flow of 5.5 A current?
| SORT You are given the half-reaction for the plating of gold, which shows the stoichiometric relationship between moles of electrons and moles of gold. You are also given the current and duration.
You must find the mass of gold that will be deposited in that time. |
GIVEN: 3 mol e^- : 1 mol Au 5.5 amps 25 min FIND: g Au |
| STRATEGIZE You need to find the amount of gold, which is related stoichiometrically to the number of electrons that have flowed through the cell. Begin with time in minutes and convert to seconds. Then, because current is a measure of charge per unit time, use the given current and the time to find the number of coulombs. Use Faraday’s constant to calculate the number of moles of electrons and the stoichiometry of the reaction to find the number of moles of gold. Finally, use the molar mass of gold to convert to mass of gold. | CONCEPTUAL PLAN
min \underset{\frac{60 s }{1 min }}{\longrightarrow }s \underset{\frac{5.5 C}{1 s }}{\longrightarrow }C \underset{ \frac{1 mol e^- }{96,485 C}}{\longrightarrow } mol e^- \underset{\frac{1 mol Au }{3 mol e ^{-}}}{\longrightarrow }mol Au \underset{\frac{196.97 g Au }{1 mol Au }}{\longrightarrow }g Au |
| SOLVE Follow the conceptual plan to solve the problem, canceling units to arrive at the mass of gold. | SOLUTION
25 \cancel{min} \times \frac{60 \cancel{s} }{1 \cancel{min} } \times \frac{5.5 \cancel{C} }{1 \cancel{s} } \times \frac{1 \cancel{mol e^-} }{96,485 \cancel{C} } \times \frac{1 \cancel{mol Au} }{3 \cancel{mol e ^{-}} } \times \frac{196.97 g Au }{1 \cancel{mol Au} }=5.6 g Au |
| CHECK The answer has the correct units (g Au). The magnitude of the answer is reasonable if you consider that 10 amps of current for 1 hour is the equivalent of about 1/3 mol of electrons (check for yourself), which would produce 1/9 mol (or about 20 g) of gold. | |