Electrolysis and Faraday's Law
Introduced| English |
|---|
| Faraday's constant/ˈfærədeɪz ˈkɒnstənt/ |
| electrolysis/ɪlekˈtrɒləsɪs/ |
Counting atoms with a stopwatch
- Pass current through a solution and metal plates out.
- The longer and stronger the current, the more metal forms.
- Charge is just electrons, and electrons build atoms.
- So a clock and an ammeter can count atoms.
Charge equals current times time
- The total charge is $Q = It$, current times time.
- More current or more time means more charge.
- That charge is a flood of electrons.
A current of $5\ \text{A}$ flows for $200\ \text{s}$. The total charge (in C)?
$Q = It = 5 \times 200 = 1000\ \text{C}$.
Faraday's constant
- Divide the charge by Faraday's constant 法拉第常数 $F$ to get moles of electrons.
- $F \approx 96{,}500\ \text{C/mol}$.
- The moles of electrons drive the electrode reaction.
To convert charge into moles of electrons, you divide by...
Moles of electrons $= Q / F$.
From electrons to product
- The half-reaction tells how many electrons each atom needs.
- Divide the moles of electrons by that number to get moles of product.
- More charge means more metal deposited.
Run an electrolysis cell
Push a current through an electrolyte and see which ions are discharged at each electrode.
Passing more charge through the cell deposits...
More charge means more electrons and more product.
Depositing copper ($\text{Cu}^{2+} + 2e^- \to \text{Cu}$) needs 2 electrons per atom. For $2\ \text{mol}$ of electrons, how much copper?
- $2\ \text{mol e}^- \div 2 = 1\ \text{mol Cu}$.
- So $1$ mole of copper plates out.
$\text{Cu}^{2+} + 2e^- \to \text{Cu}$ needs 2 electrons per atom. How many moles of Cu from $4\ \text{mol}$ of electrons?
$4\ \text{mol e}^- \div 2 = 2\ \text{mol Cu}$.
Order the steps of a Faraday's-law calculation.
Charge, then moles of electrons, then moles of product.
Depositing silver ($\text{Ag}^+ + e^- \to \text{Ag}$) needs ____ electron(s) per atom.
$\text{Ag}^+$ gains just one electron per atom.
Convert charge to moles of electrons with Faraday's constant first, then use the half-reaction's electron count to get moles of product. Watch the electrons per atom ($\text{Cu}^{2+}$ needs 2, $\text{Ag}^+$ needs 1). And $Q = It$ needs current in amperes and time in seconds.
In electrolysis 电解, the charge $Q = It$ delivers electrons. Divide by Faraday's constant ($\approx 96{,}500\ \text{C/mol}$) for moles of electrons, then divide by the half-reaction's electrons-per-atom for moles of product. More charge deposits more metal.