Electrolysis
Electrolysis电解 uses an external power source to drive a redox reaction with ΔG > 0. The applied voltage must be at least the reaction’s |E°cell|, and in practice somewhat more. For example, in the electrolysis of molten ZnCl₂:
- cathode: Zn²⁺ + 2e⁻ → Zn;
- anode: 2Cl⁻ → Cl₂ + 2e⁻;
- the overall reaction ZnCl₂ → Zn + Cl₂ has E°cell = −0.76 − 1.36 = −2.12 V, so at least 2.12 V must be applied.
Faraday’s law: charge and moles
The charge q passing through a circuit equals the current times the time (in seconds). The charge on 1 mol of electrons is the Faraday constant F:
The half-reaction stoichiometry then converts moles of electrons into moles of product: in Ag⁺ + e⁻ → Ag, each mole of electrons deposits 1 mol of Ag; in Cu²⁺ + 2e⁻ → Cu, it takes 2 mol of electrons to deposit 1 mol of Cu.
Choose a metal ion and drag the current and time to work step by step through charge → electrons → metal → mass:
Electrolysis of aqueous solutions
In an aqueous solution, water itself can react at the electrodes, so compare which species gains (or loses) electrons more easily:
- Cathode: ions with fairly large E°, such as Ag⁺ and Cu²⁺, are reduced and the metal plates out; ions such as Na⁺, K⁺, Mg²⁺ and Al³⁺ are very hard to reduce, so water is reduced to H₂ instead (2H₂O + 2e⁻ → H₂ + 2OH⁻). That’s why sodium and aluminum can only be made by electrolyzing their molten compounds.
- Anode: Br⁻, I⁻ and Cl⁻ in concentrated solution are oxidized to the halogens; SO₄²⁻ and NO₃⁻ are very hard to oxidize, so water is oxidized to O₂ instead (2H₂O → O₂ + 4H⁺ + 4e⁻).
Example 1: two electrolytic cells in series
An AgNO₃ cell and a CuSO₄ cell are connected in series, so the same charge passes through both. After some time, 1.08 g of Ag has been deposited on the cathode of the first cell. What mass of copper has been deposited on the cathode of the second? M(Ag) = 107.87 g/mol, M(Cu) = 63.55 g/mol.
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Moles of electrons from the silver. The cathode reaction is Ag⁺ + e⁻ → Ag, so each mole of Ag needs 1 mol of electrons (in mol):
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Use the series connection. In a series circuit the current is the same everywhere, so the same number of electrons passes through both cells: 0.0100 mol in the second cell too.
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Moles of copper. The cathode reaction is Cu²⁺ + 2e⁻ → Cu, so each mole of Cu needs 2 mol of electrons:
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Mass of copper.
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Check. Copper’s molar mass is about 0.59 times silver’s, and each Cu²⁺ needs 2 electrons, so the copper mass should be about 0.59 ÷ 2 ≈ 0.295 times the silver mass: 1.08 × 0.295 ≈ 0.32 g. It agrees. The current and time were never needed.
Example 2: how much aluminum per hour?
Aluminum is made industrially by electrolyzing molten Al₂O₃ dissolved in cryolite, and a single cell can carry a current of 1.00 × 10⁵ A. What is the greatest mass of aluminum, in kg, such a cell can produce per hour? M(Al) = 26.98 g/mol.
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Charge passed. q = It, with 1 h = 3600 s (in C):
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Charge → moles of electrons. Divide by the Faraday constant, F = 96 485 C/mol:
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Electrons → aluminum. The cathode reaction is Al³⁺ + 3e⁻ → Al, so 3 mol of electrons deposit 1 mol of Al:
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Mass. In g:
That is .
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Connect to practice. “Greatest” because we assume all the current goes into making aluminum (100% current efficiency). Each Al³⁺ needs 3 electrons, which is why making aluminum takes so much electricity.
- Electrolysis of CuSO₄ solution, cathode
Cu²⁺ + 2e⁻ → Cu.
- Electrolysis of molten NaCl, cathode
A molten salt contains no water, so Na⁺ is reduced.
- Electrolysis of concentrated NaCl solution, cathode
Na⁺ is very hard to reduce, so water is reduced.
- Electrolysis of molten NaCl, anode
2Cl⁻ → Cl₂ + 2e⁻.
- Electrolysis of Na₂SO₄ solution, anode
SO₄²⁻ is very hard to oxidize, so water is oxidized.
- Electrolysis of KI solution, anode
I⁻ is easily oxidized to I₂.
- Electrolysis of AgNO₃ solution, cathode
Ag⁺ + e⁻ → Ag.
- Electrolysis of Na₂SO₄ solution, cathode
Water is reduced.
Common mistakes
Mistake 1: “Metal ions in an aqueous solution always plate out during electrolysis.” Ions such as Na⁺, K⁺, Mg²⁺ and Al³⁺ are very hard to reduce in water; H₂ forms at the cathode instead.
Mistake 2: “Moles of electrons equal moles of metal.” Divide by the charge on the ion: depositing 1 mol of Cu takes 2 mol of electrons, and 1 mol of Al takes 3 mol.
Mistake 3: “Use minutes for t in q = I × t.” 1 A = 1 C/s, so the time must be in seconds.
Quick check
A current of 0.500 A flows for 10.0 min. How much charge passes?
q = 0.500 × 600 = 300 C. A doesn’t convert minutes to seconds.
How many moles of electrons are needed to deposit 0.10 mol of Al from a molten aluminum compound?
Al³⁺ + 3e⁻ → Al, so 0.10 × 3 = 0.30 mol.
An AgNO₃ cell and a CuSO₄ cell are connected in series. In the same time, what is the mole ratio of Ag to Cu deposited?
The same number of electrons passes; each Ag needs 1 electron and each Cu needs 2.
KI solution is electrolyzed with inert electrodes. What mainly forms at the cathode and the anode?
K⁺ is very hard to reduce, so water forms H₂ at the cathode; I⁻ is easily oxidized, so I₂ forms at the anode.
What is the minimum theoretical voltage for electrolyzing molten ZnCl₂ (E°: Zn²⁺/Zn −0.76 V, Cl₂/Cl⁻ +1.36 V)?
For ZnCl₂ → Zn + Cl₂, E°cell = −0.76 − 1.36 = −2.12 V, so at least 2.12 V must be applied.
To deposit twice the mass of nickel in the same time, you should:
The mass deposited is proportional to the charge q = It; for the same time, doubling the current doubles the mass.
AP-style free response
In the electrolytic refining of copper, impure copper (with small amounts of Zn, Ag and other impurities) is the anode, a thin sheet of pure copper is the cathode, and the electrolyte is a solution of CuSO₄ and H₂SO₄. E°: Ag⁺/Ag +0.80 V, Cu²⁺/Cu +0.34 V, Zn²⁺/Zn −0.76 V.
(a) Write the half-reaction for copper at the anode and the half-reaction at the cathode.
(b) With a current of 15.0 A for 8.00 h, what is the greatest mass of copper that can be deposited on the cathode? M(Cu) = 63.55 g/mol.
(c) Zinc in the impure copper also dissolves at the anode, but silver doesn’t; it falls to the bottom of the cell as “anode mud.” Explain using E° values.
(d) Zn²⁺ builds up in the solution. Why doesn’t it plate out on the cathode along with the copper?
(e) Why does this process require an external power source?
Model answer and scoring
(a) Anode: Cu → Cu²⁺ + 2e⁻; cathode: Cu²⁺ + 2e⁻ → Cu.
(b) q = 15.0 × 8.00 × 3600 = 4.32 × 10⁵ C; n(e⁻) = 4.32 × 10⁵ ÷ 96 485 = 4.48 mol; n(Cu) = 2.24 mol; m = 2.24 × 63.55 = 142 g.
(c) Zn has a smaller E° than Cu, so it is more easily oxidized and also dissolves as Zn²⁺ at the anode; Ag has a larger E° than Cu, so it is harder to oxidize, isn’t oxidized at the voltage that dissolves copper, and falls off as a solid.
(d) Zn²⁺ has a much smaller E° than Cu²⁺, so it is much harder to reduce; Cu²⁺ is reduced preferentially at the cathode, so the copper deposited is pure.
(e) The anode and cathode reactions are the same half-reaction run in opposite directions, so the overall change just moves copper from the anode to the cathode (E°cell = 0) and can’t produce a current on its own; electrical energy is needed to drive the transfer of copper (and to overcome resistance).
Scoring: (a) both half-reactions; (b) charge, electrons, moles and mass of copper; (c) E° of Zn and Ag compared with Cu; (d) Zn²⁺ harder to reduce; (e) not spontaneous, so electrical energy is needed.
Summary
- Electrolysis uses electrical energy to drive a reaction with ΔG > 0; the applied voltage must be at least |E°cell|.
- q = It (t in seconds), n(e⁻) = q/F, then use the half-reaction to convert to moles and mass of product; cells in series pass the same number of electrons.
- In aqueous solution, ions of active metals aren’t reduced (H₂ forms), and oxoanions aren’t oxidized (O₂ forms).
AP tip: In Faraday’s law calculations, show each step with units, charge → moles of electrons → moles of product → mass, using the electron coefficient from the half-reaction to convert.