Electrolysis and Faraday's law (HL)

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Using an **external power supply** to drive a **non-spontaneous** redox reaction in a molten or aqueous electrolyte.

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It is **negative**; **cations are reduced** there (gain electrons).

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It is **positive**; **anions are oxidised** there (lose electrons). Remember **PANIC** — Positive Anode Is oxidation.

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The **metal** at the cathode and the **non-metal** at the anode (e.g. molten NaCl → Na + Cl₂).

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In an **aqueous** electrolyte, **water competes** with the ions, so only **one** species is discharged at each electrode.

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**Position in the E° series**, **ion concentration**, and the **electrode material / overpotential**.

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**Hydrogen gas** — water is reduced (2H₂O + 2e⁻ → H₂ + 2OH⁻) because the metal ion is too reactive.

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**Chlorine gas** — a concentrated halide is discharged in preference to O₂.

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$Q = I\,t$ — charge (C) = current (A) × time (s). NOT in the data booklet.

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$n(e^{-}) = \dfrac{Q}{F}$ with $F = 96\,500$ C mol⁻¹ (given in the booklet).

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The charge carried by **one mole of electrons**, **F = 96 500 C mol⁻¹**.

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**Q = It → n(e⁻) = Q/F → ÷ electrons in the half-equation → moles of product → m = nM**.