Question 13.EP.6: An electrolysis cell that deposits gold (from Au^+(aq)) oper...
An electrolysis cell that deposits gold (from Au^{+}(aq)) operates for 15.0 minutes at a current of 2.30 A. What mass of gold is deposited?
Strategy
As in any stoichiometry problem we need a balanced chemical equation, so we will start by writing the half-reaction for gold reduction. To determine the mass of gold deposited, we must calculate the number of moles of electrons used from the current and the time. We can use the half-reaction to obtain a mole ratio and convert moles of electrons into moles of gold. Once we have moles of gold, we convert to mass using the molar mass, as we have done many times in stoichiometry problems.
First write the balanced half-reaction:
Au^{+}(aq) + e^{-} \longrightarrow Au(s)
Next calculate moles of electrons based on current and time:
Q = I × t = (2.30 C s^{−1})(900 s) = 2.07 × 10^{3} C
\left(2.07\times 10^{3} \ C\right)\times \left(\frac{1 \ mol \ e^{-}}{96,485 \ C} \right)=2.15\times 10^{-2}mol \ e^{-}
Now we note that the mole ratio of electrons to gold is 1:1, which means that we also have 2.15 × 10^{-2} mol of Au.
(2.15 × 10^{-2} mol Au) × (197 g mol^{-1}) = 4.23 g Au
Discussion
The latter part of this problem looks familiar, as we have been carrying out stoichiometry problems similar to this since Chapter 3. The only changes from our previous experience were the use of a half-reaction (so we can see the electrons explicitly) and the use of some physical laws that govern electricity to obtain moles of electrons in the first place.