the design and operation of electrolytic cells for the commercial production of chemicals, including comparison with galvanic cells, the polarity of electrodes in each, the difference between molten and aqueous electrolysis, and the application of Faraday's laws using Q = It and n(e) = Q/F to calculate the mass of substance produced or consumed

What this dot point is asking

VCAA wants the design and operation of electrolytic cells (used commercially to produce chemicals or refine metals), the comparison with galvanic cells (energy direction, polarity, spontaneity), the difference between molten and aqueous electrolysis in terms of the species discharged, and the quantitative application of Faraday's laws using Q = It and n(e) = Q / F to calculate the mass of substance produced or consumed.

The answer

What an electrolytic cell does

An electrolytic cell uses an external electrical energy source (a battery or DC power supply) to drive a non-spontaneous redox reaction. The cell consumes electrical energy and produces chemicals (the reverse of a galvanic cell, which produces electrical energy from a spontaneous reaction).

Components

An electrolytic cell has three components:

1. External power supply that forces electrons in the desired direction.
2. Two electrodes (often inert, like graphite or platinum, but sometimes reactive, like copper in copper plating) immersed in the same electrolyte.
3. Electrolyte: a single compartment containing a molten salt or an aqueous solution of ions.

There is no salt bridge because both electrodes share the same electrolyte.

Polarity and direction

• Cathode (reduction) is connected to the negative terminal of the power supply. Cations migrate to the cathode and gain electrons.
• Anode (oxidation) is connected to the positive terminal of the power supply. Anions migrate to the anode and lose electrons.

This is the opposite polarity to a galvanic cell, but in both cells oxidation occurs at the anode and reduction at the cathode. (AN OX, RED CAT still applies.)

Galvanic vs electrolytic: the comparison

Molten vs aqueous electrolysis

Molten electrolysis. Only the cations and anions of the salt itself are present. The cation is reduced at the cathode; the anion is oxidised at the anode.

Example. Molten NaCl:

• Cathode: Na^+ + e^- -> Na(l)
• Anode: 2Cl^- -> Cl2(g) + 2e^-

Aqueous electrolysis. Water (H2O) is also present, and it can be reduced or oxidised in competition with the dissolved ions. To predict the product, compare reduction potentials of all possible reductions at the cathode (and oxidations at the anode), and choose the most likely (typically the species with the most positive reduction potential for cathode reduction, and the most negative reduction potential when reversed for anode oxidation).

Example. Aqueous NaCl (brine):

• Cathode options: Na^+ + e^- -> Na (E° = -2.71 V) or 2H2O + 2e^- -> H2 + 2OH^- (E° = -0.83 V at standard concentrations, but reaches about -0.41 V at neutral pH). The water reduction is much more favourable, so H2 gas is produced at the cathode, not Na.
• Anode options: 2Cl^- -> Cl2 + 2e^- or 2H2O -> O2 + 4H^+ + 4e^-. Standard potentials predict O2; in practice, Cl2 is produced because of "overvoltage" effects and high [Cl^-] (the chlor-alkali process).

VCAA exam questions typically focus on molten electrolysis for clean predictions; aqueous questions may include hints or VCAA-supplied data.

Industrial examples

• Hall-Héroult process: electrolysis of molten Al2O3 dissolved in cryolite to produce aluminium metal (cathode) and oxygen at carbon anodes (which are consumed as CO2). Aluminium cannot be extracted from its ore by chemical reduction with carbon because Al is too reactive.
• Chlor-alkali process: electrolysis of brine (aqueous NaCl) to produce Cl2, H2 and NaOH.
• Electroplating: depositing a thin layer of one metal (e.g. silver, chromium, nickel) onto another by electrolysis of a solution of the plating metal's ions.
• Electrorefining of copper: a crude copper anode is dissolved, and pure Cu is deposited at the cathode from CuSO4 solution.

Faraday's laws: quantifying electrolysis

To calculate the amount of substance produced or consumed:

Q = I × t (charge in coulombs)
n(e^-) = Q / F (moles of electrons; F = 96,500 C mol^-1)
n(substance) = n(e^-) / z (z = number of electrons per mole of substance from the half-equation)
m = n × M (mass in grams)

Where:

• I = current in amperes (A; coulombs per second)
• t = time in seconds (convert from minutes or hours!)
• F = Faraday's constant = 96,500 C mol^-1 of electrons
• z = electrons per mole of substance, from the balanced half-equation

For two cells in series (sharing the same current and time), the moles of electrons is the same in both, so the moles of different substances produced are in the ratio of 1/z. (For example, in series cells producing Cu^2+ and Ag^+ deposits, moles of Ag = 2 × moles of Cu.)

Worked example

An aluminium smelter passes a current of 100,000 A through a Hall-Héroult cell for 24.0 hours. Calculate the mass of aluminium produced.

Half-equation at the cathode:
Al^3+ + 3e^- -> Al(l)

Step 1. Time in seconds: t = 24.0 × 3600 = 86,400 s.

Step 2. Charge: Q = I × t = 100,000 × 86,400 = 8.64 × 10^9 C.

Step 3. Moles of electrons: n(e^-) = Q / F = 8.64 × 10^9 / 96,500 = 8.953 × 10^4 mol.

Step 4. Moles of Al: n(Al) = n(e^-) / 3 = 2.984 × 10^4 mol.

Step 5. Mass: m = n × M = 2.984 × 10^4 × 27.0 = 806,000 g = 806 kg = 0.806 tonnes of aluminium per day per cell.

A modern smelter has hundreds of these cells operating in series, producing thousands of tonnes of aluminium per day, consuming a huge amount of electrical energy (one reason aluminium production is co-located with cheap hydropower).

Common traps

Mixing up anode and cathode polarity. In a galvanic cell, the anode is negative and the cathode is positive; in an electrolytic cell, the anode is positive and the cathode is negative. In both cells, oxidation is at the anode and reduction is at the cathode.

Forgetting to convert time to seconds. Q = It requires t in seconds. Multiply minutes by 60 or hours by 3600.

Dropping the z factor. Each species has its own electron count. Cu^2+ -> Cu needs 2 e^- per atom; Al^3+ -> Al needs 3 e^-; Ag^+ -> Ag needs 1 e^-. Forgetting this is the most common Faraday's-law error.

Predicting Na from aqueous NaCl. In aqueous solution, water is reduced at the cathode in preference to Na^+ (which is too far up the series). Molten NaCl gives Na; aqueous gives H2.

Including a salt bridge in an electrolytic cell. Electrolytic cells use a single electrolyte with no salt bridge.

Forgetting that electrolysis is forced (non-spontaneous). The external power supply provides the energy. An "electrolytic cell with no power source" is not a thing; with no power source it would either do nothing or reverse to behave as a galvanic cell.

In one sentence

An electrolytic cell uses an external power supply to drive a non-spontaneous redox reaction, with oxidation at the (positive) anode and reduction at the (negative) cathode in a single electrolyte, and the masses of products are calculated from Faraday's laws using Q = It, n(e) = Q/F and the electron count z from the half-equation.