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.)

Examples in context

Example 1. Aluminium production at Portland Aluminium smelter. Portland Aluminium electrolyses molten $\text{Al}_2 \text{O}_3$ dissolved in cryolite using the Hall-Heroult process. Cathode half-equation: $\text{Al}^{3+} + 3 e^- \to \text{Al}$, $z = 3$. Each cell draws $350{,}000 \, \text{A}$ at $4.5 \, \text{V}$. Charge per hour $Q = 350{,}000 \times 3600 = 1.26 \times 10^9 \, \text{C}$; moles of electrons $= 1.26 \times 10^9 / 96{,}500 = 1.306 \times 10^4$ mol; moles of aluminium $= 1.306 \times 10^4 / 3 = 4353$ mol; mass $= 4353 \times 27.0 = 117 \, \text{kg/hour}$ per pot. The smelter operates $360$ pots, giving $42 \, \text{tonnes/hour}$ and consuming $470 \, \text{MWh}$ daily. Power, drawn from Loy Yang, accounts for $33\%$ of operating cost.

Example 2. Chrome plating at Bendigo Mining and Engineering. A regional electroplating workshop chrome-plates truck bumpers using chromic acid baths. Cathode half-equation: $\text{CrO}_3 + 6 \text{H}^+ + 6 e^- \to \text{Cr} + 3 \text{H}_2 \text{O}$, $z = 6$. To deposit a $25 \, \mu\text{m}$ layer ($\rho_{Cr} = 7.19 \, \text{g/cm}^3$) on a $0.50 \, \text{m}^2$ bumper requires mass $= 0.50 \times 10^4 \times 0.0025 \times 7.19 = 89.9 \, \text{g}$; moles $= 89.9 / 52.0 = 1.729$ mol; moles of electrons $= 1.729 \times 6 = 10.37$ mol; charge $= 10.37 \times 96{,}500 = 1.001 \times 10^6 \, \text{C}$. At $400 \, \text{A}$, time $= 2503 \, \text{s} = 41.7 \, \text{minutes}$.

Try this

Q1. Compare a galvanic cell and an electrolytic cell. State two similarities and two differences. [4 marks]

• Cue. Similar: redox; oxidation at anode, reduction at cathode. Different: galvanic spontaneous (produces energy), electrolytic non-spontaneous (consumes); galvanic anode negative, electrolytic anode positive.

Q2. A current of $5.00 \, \text{A}$ flows through a $\text{CuSO}_4$ solution for $30.0 \, \text{minutes}$. Calculate the mass of copper deposited on the cathode. [4 marks]

• Cue. $Q = 5.00 \times 1800 = 9000 \, \text{C}$; $n(e) = 9000 / 96{,}500 = 0.0933$ mol; $\text{Cu}^{2+} + 2 e^- \to \text{Cu}$, $n(\text{Cu}) = 0.0467$ mol; mass $= 0.0467 \times 63.5 = 2.96 \, \text{g}$.

Q3. In the electrolysis of aqueous sodium chloride: (a) Write the cathode and anode half-equations. (b) Calculate the volume of $\text{H}_2$ at STP from $2 \, \text{A}$ for $1 \, \text{hour}$. (c) Explain why $\text{Na}$ is not produced. [2+2+2 marks]

• Cue. (a) Cathode: $2 \text{H}_2 \text{O} + 2 e^- \to \text{H}_2 + 2 \text{OH}^-$. Anode: $2 \text{Cl}^- \to \text{Cl}_2 + 2 e^-$. (b) $Q = 7200 \, \text{C}$; $n(e) = 0.0746$; $n(\text{H}_2) = 0.0373$; $V = 0.0373 \times 22.7 = 0.847 \, \text{L}$ at STP. (c) Water reduced preferentially; $\text{Na}^+$ has insufficient oxidising power vs $\text{H}_2 \text{O}$.