24.1 Electrolysis

At the cathode reduction occurs; at the anode oxidation occurs. In molten salts the ions present decide the products, while aqueous solutions also contain water and product selection depends on relative electrode potentials and concentration.

List all plausible half-equations, compare their ease of discharge, then balance electrons. Concentration can make a less favourable species discharge preferentially at an inert electrode.

Molten NaCl gives Na at the cathode and Cl₂ at the anode. Aqueous NaCl generally gives H₂ at the cathode and Cl₂ at the anode under concentrated brine conditions.

Do not transfer molten predictions to aqueous electrolysis without including water as a competing reactant.

The Faraday constant F is the charge carried by one mole of electrons. F = L e, where L is the Avogadro constant and e is the elementary charge.

Use coulombs consistently and remember that one electron carries charge e, while one mole carries F. This relationship connects microscopic charge to measurable electrolysis.

Using L≈6.02×10²³ mol⁻¹ and e≈1.60×10⁻¹⁹ C gives F≈9.65×10⁴ C mol⁻¹.

F is not the charge on one electron, and the symbol L here is not a current or length variable.

Charge passed is Q = It, with current I in amperes and time t in seconds. Convert charge into moles of electrons using Q/F, then use the half-equation to find mass or gas volume.

The stoichiometric coefficient matters: depositing one mole of a 2+ metal needs two moles of electrons. Keep molar mass and gas-volume conditions separate from the electrochemical calculation.

A 2.00 A current for 965 s gives Q=1930 C and about 0.0200 mol e⁻. Cu²⁺ + 2e⁻ → Cu therefore deposits 0.0100 mol Cu.

Do not use minutes in Q=It or assume one electron per ion without reading the half-equation.

An electrolytic determination measures current and time, finds the amount of substance liberated from a known half-equation, and relates charge per mole of electrons to the number of electrons and elementary charge.

Measure mass or gas volume accurately, convert to moles, use the electron ratio, and calculate F = Q/n(e⁻). Dividing F by e gives L.

If a silver cathode gains mass corresponding to 0.0100 mol Ag⁺ after 965 C, then 0.0100 mol e⁻ passed and F≈96,500 C mol⁻¹; with e known, L follows.

The measured substance amount is not automatically the electron amount; use the electrode half-equation and account for side reactions.