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24.1 Electrolysis

Syllabus
9701–2028–2029
Topic
24.1
Level
A2

Learning objectives

24.1.1The identities of substances liberated during• Predict the identities of substances liberated during electrolysis from the state of electrolyte (molten or aqueous), position in the redox series (electrode potential) and concentration24.1.2The relationship F = Le between the Faraday• State/apply: the relationship F = Le between the Faraday constant, F, the Avogadro constant, L, and the charge on the electron, e24.1.3Calculate• Calculate:: (a) the quantity of charge passed during electrolysis, using Q = I t; (b) the mass and/or volume of substance liberated during electrolysis24.1.4Determination of a value of the Avogadro• Describe the determination of a value of the Avogadro constant by an electrolytic method ⦵ ⦵

Predict electrolysis products from medium, potentials and concentration

At the cathode, cations or water are reduced; at the anode, anions, water or an active electrode are oxidised. First identify every species actually present, including water in an aqueous electrolyte, then compare plausible half-equations and apply concentration evidence.

Decision Cathode Anode
electrode process reduction: electrons on left oxidation: electrons on right
molten simple salt its cation is reduced its anion is oxidised
aqueous electrolyte, inert electrodes compare solute cation with reduction of water/H⁺ compare solute anion with oxidation of water/OH⁻
competing aqueous species electrode potential indicates thermodynamic ease; concentration can shift which species is preferentially discharged electrode potential indicates thermodynamic ease; high concentration can favour an ion that otherwise competes poorly
Electrolyte with inert electrodes Cathode product Anode product Key reason
molten NaCl Na Cl₂ only Na⁺ and Cl⁻ are present
concentrated aqueous NaCl H₂ Cl₂ water beats Na⁺ at the cathode; concentrated Cl⁻ favours chlorine at the anode
dilute aqueous NaCl H₂ O₂ water/OH⁻ oxidation becomes dominant at the anode
aqueous CuSO₄ Cu O₂ Cu²⁺ is reduced; sulfate is not preferentially oxidised

2 HX2O(l)+2 eX−→HX2(g)+2 OHX−(aq)\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}

2 ClX−(aq)→ClX2(g)+2 eX−\ce{2Cl-(aq) -> Cl2(g) + 2e-}

Do not copy molten products into an aqueous case or use a memorised ion list without checking concentration and electrode material. Electrode potentials are tabulated as reductions, so reverse the selected anode half-equation to show oxidation.

F = Le links one electron to one mole of electrons

F=LeF=Le

Symbol Meaning Approximate value / unit
F charge carried by one mole of electrons 9.65 × 10⁴ C mol⁻¹
L Avogadro constant: number of entities per mole 6.02 × 10²³ mol⁻¹
e magnitude of charge on one electron 1.60 × 10⁻¹⁹ C

F=(6.02×1023)(1.60×10−19)≈9.63×104 C mol−1F=(6.02\times10^{23})(1.60\times10^{-19})\approx9.63\times10^4\ \mathrm{C\ mol^{-1}}

Rearrange as L = F/e or e = F/L. Use the magnitude of electron charge in this counting relationship: the negative sign indicates the electron's charge direction, while F is quoted as a positive amount of charge per mole.

F is not the charge on one electron, and e is not the charge on one mole. Multiplying the per-electron charge by the number per mole is what produces C mol⁻¹.

Convert current and time into mass or gas volume through electrons

Q=It⟶n(e−)=QF→half-equationn(product)Q=It\quad\longrightarrow\quad n(e^-)=\frac{Q}{F}\quad\xrightarrow{\text{half-equation}}\quad n(\mathrm{product})

Use I in A = C s⁻¹ and t in seconds, so Q is in C. The electron-to-product ratio comes from the balanced electrode half-equation; only after finding product moles should you use molar mass or molar gas volume.

For molten MgBr₂ electrolysed at 2.20 A for 15.0 min:

Step Result
Q = It 2.20 × (15.0 × 60) = 1980 C
n(e⁻) = Q/F 1980/96500 = 0.0205 mol
Mg²⁺ + 2e⁻ → Mg n(Mg) = 0.0205/2 = 0.0103 mol
mass = nM 0.0103 × 24.3 = 0.249 g ≈ 0.25 g

For O₂ formed at 0.75 A for 35.0 min at room temperature:

Step Result
Q 0.75 × (35.0 × 60) = 1575 C
4OH⁻ → O₂ + 2H₂O + 4e⁻ n(O₂) = 1575/(4 × 96500) = 4.08 × 10⁻³ mol
V = n × 24.0 dm³ mol⁻¹ V(O₂) = 0.0979 dm³

Do not use minutes in Q = It, assume one electron per product, multiply by F when finding electron moles, or use 24.0 dm³ mol⁻¹ unless room-temperature gas conditions are appropriate.

Determine Avogadro's constant from copper electrolysis

Clean, dry and weigh copper electrodes, place them in aqueous copper(II) sulfate, pass a measured steady current for a measured time, then rinse, dry and reweigh. Use the copper mass change and Cu²⁺/Cu half-equation to find the moles of electrons associated with the measured charge.

Cu(s)→CuX2+(aq)+2 eX−\ce{Cu(s) -> Cu^{2+}(aq) + 2e-}

Example measurements: I = 0.17 A, t = 40.0 min, copper mass change = 0.13 g, Aᵣ(Cu) = 63.5 and e = 1.60 × 10⁻¹⁹ C.

Q=0.17×(40.0×60)=408 CQ=0.17\times(40.0\times60)=408\ \mathrm{C}

n(e−)=2(0.1363.5)=4.09×10−3 moln(e^-)=2\left(\frac{0.13}{63.5}\right)=4.09\times10^{-3}\ \mathrm{mol}

F=Qn(e−)≈4084.09×10−3=9.96×104 C mol−1F=\frac{Q}{n(e^-)}\approx\frac{408}{4.09\times10^{-3}}=9.96\times10^4\ \mathrm{C\ mol^{-1}}

L=Fe=9.96×1041.60×10−19≈6.23×1023 mol−1L=\frac{F}{e}=\frac{9.96\times10^4}{1.60\times10^{-19}}\approx6.23\times10^{23}\ \mathrm{mol^{-1}}

Rinsing and drying before weighing, measuring current throughout, timing accurately and limiting side reactions improve the result. The measured copper amount is converted to electron amount with the factor 2; it is not automatically equal to n(e⁻).