Electrolytic cells
In an electrolytic cell, electrical energy from outside makes a reaction happen that would not happen by itself (ΔG > 0). This is electrolysis.
- The cathode is joined to the negative end of the battery. Cations go there and are reduced.
- The anode is joined to the positive end. Anions go there and are oxidised.
| Galvanic cell | Electrolytic cell | |
|---|---|---|
| Energy | chemical → electrical | electrical → chemical |
| Reaction | spontaneous (ΔG < 0) | non-spontaneous (ΔG > 0) |
| Anode sign | − | + |
| Cathode sign | + | − |
Uses: getting reactive metals (Na, Mg, Al), refining copper, electroplating, and making NaOH and Cl₂.
Faraday's laws of electrolysis
Michael Faraday measured how much substance changes during electrolysis.
First law
The mass changed at an electrode is proportional to the charge passed: m = Z × Q = Z × I × t. Z is the electrochemical equivalent (grams per coulomb).
Second law
If the same charge passes through different electrolytes, the masses changed are in the ratio of their equivalent weights (molar mass ÷ electrons per ion).
How to calculate
One mole of electrons carries 96 500 C = 1 faraday (F). For Mⁿ⁺ + ne⁻ → M, you need n F to make 1 mole of M.
moles of e⁻ = I t / F, moles of M = I t / (nF), m = M × I t / (nF).
Products of electrolysis
The product depends on what is being electrolysed, what the electrodes are made of, and the concentration.
- Molten NaCl: only Na⁺ and Cl⁻ are present. Na metal at the cathode, Cl₂ at the anode.
- Aqueous NaCl: water competes. At the cathode, H₂O (E° = −0.83 V) is reduced much more easily than Na⁺ (E° = −2.71 V), so H₂ forms and the solution becomes NaOH. At the anode, water should be easier to oxidise by E°, but it needs a large extra voltage (overpotential), so Cl₂ comes out.
- Aqueous CuSO₄ with copper electrodes: Cu is plated at the cathode, and the copper anode dissolves (Cu → Cu²⁺). This is how copper is refined.
- Aqueous CuSO₄ with platinum electrodes (inert): Cu at the cathode, O₂ at the anode.
- Dilute H₂SO₄: H₂ at the cathode, O₂ at the anode. Concentrated H₂SO₄ can give peroxodisulphate (S₂O₈²⁻) at the anode.
Rule of thumb: at the cathode, the species with the higher reduction potential is reduced first; at the anode, the one that is most easily oxidised (after allowing for overpotential) reacts first.
Batteries: primary and secondary
A battery is one or more galvanic cells put together to give a steady voltage.
Primary batteries (use once)
The reaction cannot be reversed easily.
- Dry cell (Leclanché cell): zinc can = anode, carbon (graphite) rod = cathode with MnO₂ + carbon, and a paste of NH₄Cl and ZnCl₂. Anode: Zn → Zn²⁺ + 2e⁻. Cathode: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃. About 1.5 V.
- Mercury cell (watches, hearing aids): Zn–Hg amalgam anode, HgO + carbon cathode, KOH + ZnO paste. Overall Zn(Hg) + HgO → ZnO + Hg. About 1.35 V, and it stays steady because no ions change in concentration.
Secondary batteries (rechargeable)
- Lead storage battery: anode Pb, cathode grid packed with PbO₂, electrolyte ~38% H₂SO₄.
Discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Charging reverses it. Acid gets weaker as it discharges. - Nickel–cadmium cell: longer life than lead cells. Cd + 2Ni(OH)₃ → CdO + 2Ni(OH)₂ + H₂O.
Fuel cells
A fuel cell is a galvanic cell that is fed its reactants all the time, so it does not run down.
H₂–O₂ fuel cell: H₂ and O₂ bubble through porous carbon electrodes (with a Pt or Pd catalyst) into concentrated NaOH or KOH.
- Anode: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻
- Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻
- Overall: 2H₂ + O₂ → 2H₂O (about 1.23 V)
Why it matters: efficiency of about 70% (a power plant burning fuel gives about 40%), no pollution (only water), and the water is drinkable. Used in space vehicles and some buses.
Corrosion
Corrosion is the slow eating away of a metal by air, water and other chemicals around it. Rusting of iron, green layer on copper, and black layer on silver are examples.
Rusting is an electrochemical cell
- Anode (a spot on the iron): 2Fe → 2Fe²⁺ + 4e⁻ (E° = −0.44 V)
- Cathode (another spot, where air reaches the water): O₂ + 4H⁺ + 4e⁻ → 2H₂O (E° = +1.23 V). H⁺ comes from CO₂ dissolved in water.
- Overall E° = 1.67 V, so it is spontaneous.
- Fe²⁺ is then oxidised further by air to hydrated iron(III) oxide, Fe₂O₃·xH₂O = rust.
Salt water speeds it up because it has more ions and conducts better.
Prevention
- Barrier: paint, oil, grease, plastic coat.
- Galvanising: coat iron with zinc. Zinc is more active, so it corrodes first and protects iron even when scratched.
- Sacrificial anode: attach Mg or Zn blocks to ships and pipes.
- Anti-rust solutions such as bisphenol, and alloying (stainless steel).
Try it: copper plating a key
With an adult: dissolve a spoon of copper sulphate in water. Join a copper strip to the + end of a 1.5 V cell and a clean steel key to the − end, and dip both in. In 10 minutes the key turns pink with copper. Predict first with the 3D (step 1): if you double the current, how much more copper? Double the time? Also leave two iron nails in plain water and salt water for a week and compare the rust.
Key formulas and definitions
- Q = I × t (coulomb = ampere × second)
- 1 F = 96 500 C mol⁻¹ (charge on 1 mol e⁻)
- m = Z I t (Faraday's first law)
- m = M × I × t / (n × F)
- m₁/m₂ = E₁/E₂ (equivalent weights, second law)
- Lead cell: Pb + PbO₂ + 2H₂SO₄ ⇌ 2PbSO₄ + 2H₂O
- Fuel cell: 2H₂ + O₂ → 2H₂O
- Rusting: 2Fe + O₂ + 4H⁺ → 2Fe²⁺ + 2H₂O ; E° = 1.67 V
Worked examples
1. A current of 2 A flows through CuSO₄ solution for 965 s. Find the mass of copper deposited (Cu = 63.5 g mol⁻¹).
Step 1: Q = I t = 2 × 965 = 1930 C. Step 2: moles of e⁻ = 1930 / 96 500 = 0.02 mol. Step 3: Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 0.02 / 2 = 0.01 mol. Step 4: mass = 0.01 × 63.5 = 0.635 g.
2. How long must 2 A flow to deposit 1.27 g of copper?
Step 1: moles of Cu = 1.27 / 63.5 = 0.02 mol. Step 2: moles of e⁻ = 2 × 0.02 = 0.04 mol. Step 3: Q = 0.04 × 96 500 = 3860 C. Step 4: t = Q / I = 3860 / 2 = 1930 s (about 32 min).
3. A current of 0.5 A is passed through AgNO₃ solution for 1930 s. Find the mass of silver deposited (Ag = 108).
Step 1: Q = 0.5 × 1930 = 965 C. Step 2: moles of e⁻ = 965 / 96 500 = 0.01 mol. Step 3: Ag⁺ + e⁻ → Ag, so 0.01 mol Ag. Step 4: mass = 0.01 × 108 = 1.08 g.
4. How many coulombs are needed to make 1 mol of Al from Al³⁺?
Step 1: Al³⁺ + 3e⁻ → Al. Step 2: 1 mol Al needs 3 mol e⁻ = 3 F. Step 3: Q = 3 × 96 500 = 289 500 C.
5. The same charge passes through AgNO₃ and CuSO₄ cells in series. 1.08 g of Ag is deposited. How much Cu is deposited?
Step 1: moles of Ag = 1.08 / 108 = 0.01 mol, so 0.01 mol e⁻ passed. Step 2: Cu needs 2e⁻ per atom: moles of Cu = 0.01 / 2 = 0.005 mol. Step 3: mass of Cu = 0.005 × 63.5 = 0.3175 g.
6. How many electrons flow when 1 A passes for 60 s? (e = 1.602 × 10⁻¹⁹ C)
Step 1: Q = 1 × 60 = 60 C. Step 2: number = Q / e = 60 / (1.602 × 10⁻¹⁹). Step 3: ≈ 3.75 × 10²⁰ electrons.
7. Find the volume of H₂ at STP given by 9650 C during electrolysis of water.
Step 1: moles of e⁻ = 9650 / 96 500 = 0.1 mol. Step 2: 2H⁺ + 2e⁻ → H₂, so moles of H₂ = 0.05 mol. Step 3: volume = 0.05 × 22.4 L = 1.12 L.
8. Why is H₂, not Na, formed at the cathode when aqueous NaCl is electrolysed?
Step 1: Compare: Na⁺ + e⁻ → Na, E° = −2.71 V; 2H₂O + 2e⁻ → H₂ + 2OH⁻, E° = −0.83 V. Step 2: The higher (less negative) value is reduced first. Step 3: So water is reduced and H₂ is given off; Na⁺ stays in solution as NaOH.
Common mistakes
- Mixing up signs: in electrolysis the anode is + and the cathode is −, opposite to a galvanic cell. The anode is still where oxidation happens.
- Forgetting n: Cu²⁺ needs 2 mol e⁻ per mol Cu, Al³⁺ needs 3. Divide moles of electrons by n.
- Using minutes instead of seconds in Q = I × t.
- Thinking aqueous NaCl gives Na at the cathode. Water is reduced first, giving H₂.