Electrolysis is stoichiometry with electrons as a reactant. The current tells you how many electrons pass, and the half-reaction tells you how many each ion needs. From there, moles to grams as usual. This builds on electrolytic cells.
Key idea
Charge $q = It$ (coulombs = amperes × seconds). Moles of electrons $= q/F$, with $F = 96\,485\ \text{C/mol}$. Moles of product = moles of electrons ÷ electrons per formula (from the half-reaction). Mass = moles × molar mass.
Why it works
An ampere is one coulomb per second, so current times time is total charge. One mole of electrons carries $96\,485\ \text{C}$ (Faraday's constant, $N_A$ times the charge of one electron). Each $\mathrm{Cu^{2+}}$ needs two electrons to become a copper atom, so a mole of electrons makes only half a mole of copper. Michael Faraday found this pattern by experiment in the 1830s: the same charge always deposits amounts of different metals in proportion to molar mass divided by ionic charge.
Worked examples
Worked example: Copper plating
A current of $2.00\ \text{A}$ runs for $30.0\ \text{min}$ through $\mathrm{CuSO_4}$ solution. What mass of copper plates out?
- $q = (2.00\ \text{A})(1800\ \text{s}) = 3600\ \text{C}$.
- Moles of electrons: $\dfrac{3600}{96\,485} = 0.03731\ \text{mol}$.
- $\mathrm{Cu^{2+} + 2e^- \to Cu}$: moles of Cu $= \dfrac{0.03731}{2} = 0.01866\ \text{mol}$.
- Mass $= (0.01866)(63.55\ \text{g/mol}) = 1.19\ \text{g}$.
Answer: $1.19\ \text{g}$.
Worked example: How long?
How long must $0.500\ \text{A}$ run to plate $1.00\ \text{g}$ of silver?
- Moles of Ag $= \dfrac{1.00}{107.87} = 9.27 \times 10^{-3}\ \text{mol}$. $\mathrm{Ag^+ + e^- \to Ag}$: one electron each, so $9.27 \times 10^{-3}\ \text{mol}$ of electrons.
- $q = (9.27 \times 10^{-3})(96\,485) = 894\ \text{C}$.
- $t = \dfrac{894\ \text{C}}{0.500\ \text{A}} = 1.79 \times 10^3\ \text{s}$, about $29.8\ \text{min}$.
Answer: About 29.8 min.
Worked example: Aluminum on an industrial scale
How much charge is needed to make $1.00\ \text{kg}$ of aluminum from $\mathrm{Al^{3+}}$?
- Moles of Al $= \dfrac{1000}{26.98} = 37.1\ \text{mol}$; three electrons each: $111\ \text{mol}$ of electrons.
- $q = (111)(96\,485) = 1.07 \times 10^{7}\ \text{C}$. That is why aluminum smelters sit near cheap electricity.
Answer: $1.07 \times 10^{7}\ \text{C}$.
Battery capacity is often given in ampere-hours, a unit of charge: $1\ \text{A·h} = 3600\ \text{C}$. A phone battery rated $3.0\ \text{A·h}$ can deliver $10\,800\ \text{C}$, about $0.11\ \text{mol}$ of electrons, before it needs recharging.
In practice some current is lost to side reactions, such as making hydrogen at the cathode, so real yields are slightly below the Faraday's-law amount. Industry reports this as the current efficiency.
Common mistake
Forgetting how many electrons each ion needs, which predicts twice as much copper as can form. Check the half-reaction every time: $\mathrm{Ag^+}$ takes 1, $\mathrm{Cu^{2+}}$ takes 2, $\mathrm{Al^{3+}}$ takes 3.