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24. Electrochemistry A2

Syllabus
9701–2028–2029
Section
24
Level
A2

24.1 Electrolysis

Syllabus
9701–2028–2029
Topic
24.1
Level
A2

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⁻).

24.2 Standard electrode potentials E ⦵ , standard cell potentials E ⦵ cell and the Nernst

Syllabus
9701–2028–2029
Topic
24.2
Level
A2

Define standard electrode and standard cell potentials

A standard electrode (reduction) potential, E°, is the potential of a half-cell relative to the standard hydrogen electrode, measured under standard conditions with the half-equation written as a reduction.

A standard cell potential, E°cell, is the potential difference between two standard half-cells. It is calculated from their reduction potentials.

Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}

Use 298 K, aqueous ion concentrations of 1.00 mol dm⁻³ and standard gas pressure. Potentials are measured in volts and cannot be measured for an isolated half-cell.

E° is an intensive quantity: balance electrons in half-equations but never multiply an E° value by a stoichiometric coefficient.

The standard hydrogen electrode fixes the zero reference

2 HX+(aq)+2 eX−⇌HX2(g)E∘=0.00 V\ce{2H+(aq) + 2e- <=> H2(g)}\qquad E^\circ=0.00\ \mathrm{V}

SHE feature Standard requirement and purpose
hydrogen gas supplied at standard pressure and in contact with the electrode
H⁺(aq) 1.00 mol dm⁻³
temperature 298 K
platinised platinum inert electrical conductor and catalytic surface; not consumed

Connect the SHE to the unknown half-cell with a salt bridge and a high-resistance voltmeter. The meter polarity and cell voltage locate the unknown reduction potential relative to 0.00 V.

The platinum is not a hydrogen reactant and the SHE is not simply an acid beaker: gas pressure, concentration, temperature and gas–solution–platinum contact all matter.

Measure E° for metal, non-metal and ion–ion half-cells

Redox pair Half-cell construction Why the electrode works
metal/metal ion, e.g. Zn²⁺/Zn Zn(s) dipped in 1.00 mol dm⁻³ Zn²⁺(aq) the metal is both reactant and conductor
non-metal/non-metal ion, e.g. Br₂/Br⁻ inert Pt contacts both Br₂ and 1.00 mol dm⁻³ Br⁻ Pt conducts because no suitable solid metal conductor belongs to the pair
same element in two aqueous oxidation states, e.g. Fe³⁺/Fe²⁺ inert Pt dipped into a solution containing both ions at 1.00 mol dm⁻³ Pt transfers electrons without entering the redox equation

Maintain standard conditions, connect the test half-cell to the SHE by a non-reacting salt bridge and high-resistance voltmeter, then record both voltage and polarity. Write both reduction half-equations; use the SHE value 0.00 V and polarity to assign the sign of the test E°.

The salt bridge completes the internal circuit by ion migration and maintains electrical neutrality. Electrons travel through the external wire and voltmeter, not through the bridge.

Include every aqueous species required by the half-equation. An ion–ion half-cell involving H⁺, such as MnO₄⁻/Mn²⁺, also requires the specified standard H⁺ concentration.

Calculate E°cell by subtracting two reduction potentials

Reduction half-equation E° / V Role in the feasible cell
Cu²⁺ + 2e⁻ ⇌ Cu +0.34 more positive: reduction at cathode
Zn²⁺ + 2e⁻ ⇌ Zn −0.76 less positive: reverse for oxidation at anode

Ecell∘=Ecathode∘−Eanode∘=+0.34−(−0.76)=+1.10 VE^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=+0.34-(-0.76)=+1.10\ \mathrm{V}

Keep both tabulated numbers as reduction potentials. Select the more positive value for the cathode and subtract the less positive anode value. Balance the reaction separately; multiplying a half-equation never scales its potential.

Do not add two tabulated reduction potentials blindly. The equivalent oxidation-potential method changes the sign only when the half-equation is reversed.

Use E°cell for polarity, electron flow and feasibility

Feature in a simple galvanic cell More positive E° half-cell Less positive E° half-cell
reaction reduction oxidation
electrode name cathode anode
polarity positive negative
external electron flow receives electrons supplies electrons

Electrons flow through the external circuit from the negative anode to the positive cathode. The salt bridge carries ions to preserve charge balance; it does not carry the electrons between electrodes.

For the reaction direction used to calculate the cell, E°cell > 0 predicts thermodynamic feasibility under standard conditions. E°cell < 0 means that written direction is not feasible under those conditions; reversing it changes the sign.

Positive E°cell does not mean fast reaction. Electrode potentials predict thermodynamic direction, while activation energy controls rate.

Rank oxidising and reducing agents from reduction potentials

In a reduction half-equation, the species on the left accepts electrons and is the oxidising agent; the species on the right can donate electrons in the reverse direction and is the reducing agent.

E° trend Oxidised form on left Reduced form on right
more positive stronger oxidising agent; more readily reduced weaker reducing agent
more negative weaker oxidising agent stronger reducing agent; more readily oxidised

Because E°(Cl₂/Cl⁻) is more positive than E°(I₂/I⁻), Cl₂ is the stronger oxidising agent and I⁻ is the stronger reducing agent. Cl₂ can therefore oxidise I⁻ under standard conditions.

Compare the paired forms in correctly written reduction half-equations. A large positive E° ranks the left-hand oxidised species, not every species named in that row.

Construct a redox equation by cancelling electrons

Reduction half-equation E° / V Selected direction
Fe³⁺ + e⁻ ⇌ Fe²⁺ +0.77 forward reduction
I₂ + 2e⁻ ⇌ 2I⁻ +0.54 reverse oxidation

2 IX−(aq)→IX2(aq)+2 eX−\ce{2I-(aq) -> I2(aq) + 2e-}

2 FeX3+(aq)+2 eX−→2 FeX2+(aq)\ce{2Fe^{3+}(aq) + 2e- -> 2Fe^{2+}(aq)}

2 FeX3+(aq)+2 IX−(aq)→2 FeX2+(aq)+IX2(aq)\ce{2Fe^{3+}(aq) + 2I-(aq) -> 2Fe^{2+}(aq) + I2(aq)}

Choose the more positive reduction, reverse the other half-equation, multiply equations until electron numbers match, add, cancel electrons and any identical species, then verify both atoms and total charge.

Multiply half-equation coefficients but not E° values. No electrons may remain in the final redox equation.

Predict concentration effects from the reduction equilibrium

Ox(aq)+ze−⇌Red(aq)\mathrm{Ox(aq)}+ze^-\rightleftharpoons\mathrm{Red(aq)}

Concentration change at fixed temperature Equilibrium response Effect on reduction potential E
increase aqueous oxidised species favours reduction/right more positive
decrease aqueous oxidised species favours oxidation/left less positive
increase aqueous reduced species favours oxidation/left less positive
decrease aqueous reduced species favours reduction/right more positive

For Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), increasing [Cu²⁺] makes E more positive. Changing the amount of pure Cu(s) does not change E while solid copper remains present because its activity is constant.

Do not say every concentration increase raises E. First identify whether the changed aqueous ion is the oxidised or reduced species; omit pure solids from the concentration comparison.

Calculate a non-standard electrode potential with the Nernst equation

E=E∘+0.059zlog⁡10([oxidised species][reduced species])at 298 KE=E^\circ+\frac{0.059}{z}\log_{10}\left(\frac{[\mathrm{oxidised\ species}]}{[\mathrm{reduced\ species}]}\right)\quad\text{at 298 K}

z is the number of electrons in the written reduction half-equation. In the syllabus form, use aqueous-ion concentrations; a pure solid has unit activity and is omitted from the ratio.

For Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), E° = +0.34 V and [Cu²⁺] = 1.00 × 10⁻³ mol dm⁻³. The reduced species Cu(s) is omitted, so the ratio is 1.00 × 10⁻³ and z = 2.

E=0.34+0.0592log⁡10(1.00×10−3)=0.34−0.0885≈+0.25 VE=0.34+\frac{0.059}{2}\log_{10}(1.00\times10^{-3})=0.34-0.0885\approx+0.25\ \mathrm{V}

Diluting Cu²⁺ makes its reduction less favourable, so E should be below +0.34 V; the numerical result agrees with the qualitative prediction. For Fe³⁺/Fe²⁺, both aqueous concentrations remain in [Fe³⁺]/[Fe²⁺] and z = 1.

Use base-10 log, the oxidised/reduced order and the half-equation electron number—not an overall-cell coefficient. This 0.059 form is for 298 K.

Link standard cell potential to standard Gibbs energy

ΔG∘=−nFEcell∘\Delta G^\circ=-nFE^\circ_{cell}

Symbol Meaning and unit
n moles of electrons transferred per mole of the balanced overall reaction
F 96500 C mol⁻¹
E°cell standard cell potential in V = J C⁻¹
ΔG° standard Gibbs energy change, initially obtained in J mol⁻¹

For 2Fe³⁺(aq) + Cu(s) → 2Fe²⁺(aq) + Cu²⁺(aq), E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Cu²⁺/Cu) = +0.34 V. Thus E°cell = +0.43 V and the balanced reaction transfers n = 2 electrons.

ΔG∘=−2(96500)(+0.43)=−82990 J mol−1≈−83.0 kJ mol−1\Delta G^\circ=-2(96500)(+0.43)=-82990\ \mathrm{J\ mol^{-1}}\approx-83.0\ \mathrm{kJ\ mol^{-1}}

A positive E°cell gives a negative ΔG°, so the written reaction is thermodynamically feasible under standard conditions. Reversing the reaction reverses both signs.

n comes from electrons cancelled in the balanced overall equation. Do not omit n, multiply E° by coefficients, or report the joule result as kilojoules without dividing by 1000.