A2 Chemistry Electrochemistry Notes

V e⁻ anode · oxidation cathode · reduction

CAMBRIDGEINTERNATIONAL·CAIE

A Level Chemistry 9701 · Paper 4

A2REVISIONNOTES·PHYSICALCHEMISTRY

Electrochemistry

Complete topic notes — typed, colour-coded and worked through, with the original handwritten class pages reproduced alongside every section.

Redox & Oxidation States Balancing Half-Equations Disproportionation

Standard Electrode Potential The S.H.E. Electrochemical Cells

E°cell & Feasibility Chain Reactions Nernst Equation

Homogeneous Catalysis Electrolysis Faraday Constant

PREPARED&TAUGHTBY

Sir Fahad H. Ahmad

CHEMISTRY LEAD · MEGA LECTURE

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CAIEALEVELCHEMISTRY·9701

A2 Revision Notes

TOPIC·PHYSICALCHEMISTRY

Electrochemistry

Redox & oxidation states · standard electrode potentials · electrochemical cells · the Nernst equation · homogeneous catalysis · electrolysis · the Faraday constant. Typed notes with the original handwritten class pages alongside.

Fahad H. Ahmad

CHEMISTRY LEAD · MEGA LECTURE

World Distinction in A Levels · Top in Pakistan in O Levels · 4× Best Across 3 A Levels.

Chemistry lectures with 10M+ views worldwide.

megalecture.com WhatsApp +92 336 7801123 fahad.h.ahmad@gmail.com youtube.com/MegaLecture

CONTENTS

Redox Fundamentals

Oxidation States

Balancing Redox Equations

Disproportionation

Standard Electrode Potential

Electrode Types & the S.H.E.

The Complete Cell

Building Redox Equations from E°

Chain Reactions (Excess Reagent)

Non-Standard Conditions & Nernst

Homogeneous Catalysis

Electrolysis

The Faraday Constant

Exam Checklist

Redox Fundamentals

DEFINITIONS

Reduction — gain of electrons; oxidation state decreases (becomes more negative).

Oxidation — loss of electrons; oxidation state increases (becomes more positive).

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Direction of change on the oxidation-state number line.

Na — loses 6e− → Na+7 + 6e− oxidation

2e− + Al3+ — gains 2e− → Al+1 reduction

Oxidation States

DEFINITION

The hypothetical or actual charge on an element if the compound is considered to be ionic.

In ionic compounds the charge is real: in NaCl, Na is +1 and Cl is −1. In covalent molecules the oxidation state is assigned by electronegativity — the more electronegative atom in each bond is given both electrons.

−3 −2 −1 +1 +2 +3 reduction — gain e⁻ — more negative oxidation — lose e⁻ — more positive

ORIGINAL CL ASS NOTES — REDOX DEFINITIONS & NUMBER LINE

H—Cl H = +1, Cl = −1 Cl is more electronegative, so it takes the shared pair

Worked assignments from bonding

SPECIES ASSIGNMENT REASONING

H2S H = +1, S = −2 S more electronegative than H in both bonds

O=C=O C = +4, O = −2 each All four bonding pairs pulled to O

Cl—Cl both 0 Identical atoms — pair shared equally

H2O2 H = +1, O = −1 each O—O bond splits evenly, so O is only −1

CH3COOH CH3 carbon = −3, COOH carbon = +3 Assign bond by bond; overall sum = 0

OXIDATION STATE DEFINED · IONIC VS COVALENT ASSIGNMENT

THERULES—LEARNTHESE

Oxygen = −2, except with fluorine, or in peroxides (H2O2) where it bonds to another O.

Hydrogen = +1 with non-metals; −1 with metals (e.g. Na+1H−1, since Na 0.9 < H 2.1 on electronegativity).

Group 1 = +1, Group 2 = +2, Group 3 = +3.

Fluorine is always −1.

A neutral element = 0.

Transition metals show variable oxidation states — always calculate.

DOT-AND-CROSS STRUCTURES WITH ELECTRONS ASSIGNED BY ELECTRONEGATIVITY

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Calculating an unknown state

Set the sum of oxidation states equal to the overall charge and solve for x.

SPECIES EQUATION RESULT

NaClO3 (+1) + x + (−2 × 3) = 0 Cl = +5

HNO3 (+1) + x + (−2 × 3) = 0 N = +5

SO42− x + (−2 × 4) = −2 S = +6

NH4+ x + (+1 × 4) = +1 N = −3

NO3− x + (−2 × 3) = −1 N = +5

CH3COOH 2x + 3 − 4 + 1 = 0 Cavg = 0

Fe3O4 3x − (2 × 4) = 0 Fe = +8/3

WHYAFRACTIONALANSWER?

Fe3O4 gives Fe = +8/3 because it is really a mixed oxide — FeO · Fe2O3, containing Fe(+2) and two

Fe(+3). The +8/3 is only an average. For polyatomic salts such as NH4NO3, split into ions first (NH4+ and NO3−) and treat each separately.

THE RULE BOX, EXACTLY AS WRITTEN IN CL ASS

Balancing Redox Equations

GOVERNINGPRINCIPLE

electrons gained = electrons lost Example — Na + Al3+

Al3+ gains 3e−; each Na loses 1e−, so three Na are needed.

3 Na + Al3+ → 3 Na+ + Al

FULL WORKED CALCUL ATIONS, INCLUDING THE FE 3O 4 TRICK

Example — ClO4− + I−

Cl goes +7 → −1 (gains 8e−); I goes −1 → +5 (loses 6e−). LCM of 8 and 6 is 24, so ×3 and ×4.

3 ClO4− + 4 I− → 3 Cl− + 4 IO3−

Example — Al + Fe2+, via half-equations

oxidation 2 × ( Al → Al3+ + 3e− ) reduction 3 × ( Fe2+ + 2e− → Fe )

2 Al + 3 Fe2+ → 2 Al3+ + 3 Fe

THE ELECTRON-BOOKKEEPING ARROWS

FINDING THE MULTIPLIERS FROM ELECTRONS GAINED AND LOST

A L G E B R A O F H A L F- E Q UAT I O N S a × oxidation + b × reduction = overall redox equation

Rearranged: overall − (a × oxidation) = b × reduction — useful when a question gives you the overall equation and one half-equation, and asks for the other.

Acidic-medium example — MnO4− + C2O42−

Mn: +7 → +2 (gains 5e−). C: +3 → +4 (loses 1e− per C, so 2e− per C2O42−).

reduction ( MnO4− + 8H+ + 5e− → Mn2+ + 4H2O ) × 2 oxidation ( C2O42− → 2CO2 + 2e− ) × 5

2 MnO4− + 5 C2O42− + 16 H+ → 2 Mn2+ + 10 CO2 + 8 H2O

HALF-EQUATIONS SCALED AND ADDED

THE REARRANGEMENT, IN THE ORIGINAL HAND

Example — IO3− + N2O

2 IO3− + 5 N2O + 2 H+ → I2 + 10 NO + H2O

I: +5 → 0 (gains 5e−, ×2). N: +1 → +2 (loses 1e−, ×10).

Disproportionation

DEFINITION

The same element is simultaneously oxidised and reduced in one reaction.

MANGANATE(VII) / ETHANEDIOATE — THE CL ASSIC TITRATION REDOX

IODATE AND DINITROGEN OXIDE

3 Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3 H2O

Cl 0 → −1 in NaCl — gains 1e−, ×5

Cl 0 → +5 in NaClO3 — loses 5e−, ×1

Check on NaClO3: (+1) + x + (−2 × 3) = 0 → x = +5. ✓

Standard Electrode Potential, E°

DEFINITION

The potential difference measured between a standard electrode and a standard hydrogen electrode.

Standard conditions: 298 K, 1 atm, all solutions 1 mol dm−3.

The core interpretation

ELECTRODE POTENTIAL TENDENCY ROLE IN A CELL

Higher / more positive E° Wants to gain electrons Is reduced — forward direction

Lower / more negative E° Wants to lose electrons Is oxidised — backward direction

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CHLORINE DISPROPORTIONATING IN HOT ALKALI

THE RULE IN ONE LINE — WORTH MEMORISING

Na+ + e− ⇌ Na E° = −2.71 V very negative → Na strongly wants to lose e−

Cl2 + 2e− ⇌ 2Cl− E° = +1.36 V very positive → Cl2 strongly wants to gain e−

DEFINITION OF E° AND THE NA | NA⁺ METAL / METAL-ION ELECTRODE

WAT C H T H E S I G N C O N V E N T I O N

All electrode potentials are quoted as reduction potentials — always written left-to-right as gaining electrons. A negative E° simply means the reverse (oxidation) is favoured.

Electrode Types & the S.H.E.

TYPE CONSTRUCTION EXAMPLE

Metal / metal ion

Metal rod dipped into a solution of its own ion Na | Na+ E° = −2.71 V

Gas / aqueous ion

Inert Pt electrode, gas bubbled at 1 atm over it, in 1 mol dm−3 ion solution

Pt | Cl2 | Cl− E° =

+1.36 V

Ion / ion Inert Pt electrode in a solution containing both oxidation states Pt | Fe3+, Fe2+ E° =

+0.77 V

GAS / AQUEOUS-ION ELECTRODE — CL 2 OVER A PT ELECTRODE

REFERENCEELECTRODE—S.H.E.

The Standard Hydrogen Electrode is assigned E° = 0.00 V by definition. Every other electrode potential is measured against it.

2 H+(aq) + 2e− ⇌ H2(g) E° = 0.00 V

Set-up: Pt electrode, H2 gas at 1 atm, H+(aq) = 1 mol dm−3, 298 K.

ION / ION ELECTRODE — FE 3 +, FE 2 + ON PT

The Standard Hydrogen Electrode — the universal reference, E° = 0.00 V.

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←H₂ (g), 1 atm H⁺ H⁺ H⁺ (aq) = 1 mol dm⁻³ E° = 0.00 V Pt

The Complete Cell

Connecting two half-cells — one that loses electrons and one that gains them — produces electrical energy.

T H R E E R U L E S T H AT D E C I D E E V E RY T H I N G

Electrons travel through the wire towards the higher E°.

The higher-E° half-cell is reduced; the lower-E° half-cell is oxidised.

The salt bridge balances the charges in the solutions and completes the circuit.

S.H.E. AS DRAWN IN CL ASS

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A complete cell: electrons flow externally to the higher-E° electrode; the salt bridge carries ions to balance charge.

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e⁻ flow →towards the higher E° salt bridge higher E° · gains e⁻ · REDUCED lower E° · loses e⁻ · OXIDISED

Cl₂ + 2e⁻ →2Cl⁻ E° = +1.36 V H₂ →2H⁺ + 2e⁻ E° = 0.00 V

Example — Fe3+/Fe2+ against O2/OH−

reduction 4 Fe3+ + 4e− → 4 Fe2+ E° = +0.77 V oxidation 4 OH− → O2 + 2H2O + 4e− E° = +0.40 V

4 Fe3+ + 4 OH− → 4 Fe2+ + O2 + 2 H2O

THE FULL CELL DIAGRAM FROM CL ASS — CL 2/CL − AGAINST THE S.H.E.

CELLPOTENTIAL

E°cell = E°(reduction, higher) − E°(oxidation, lower)

0.77 − 0.40 = +0.37 V

E°cell > 0 → the reaction is feasible.

OXYGEN ELECTRODE VS IRON ELECTRODE, AND THE E° C E L L FEASIBILITY TEST

Building Redox Equations from E° Data

When a question gives you only the substances — no electrode reactions — follow this routine.

W R I T I N G T H E E Q UAT I O N D I R E C T LY

Take the two half-equations, run the higher-E° one forwards and the lower-E° one backwards, scale so electrons cancel, then read reactants → products.

e.g. 2Na+ + 2e− ⇌ 2Na (−2.71 V) and I2 + 2e− ⇌ 2I− (+0.54 V) give 2Na + I2 → 2Na+ + 2I−

Identify every species actually present, including spectator ions from the salts.

List all electrode reactions from the data booklet that involve those species.

Highest E° gains electrons (forward direction); lowest E° loses electrons (reverse direction).

Check the species you selected are ones you actually have. You cannot oxidise something that is already fully oxidised, or reduce something absent from the mixture.

Balance electrons, then combine into the overall equation and compute E°cell.

READING REACTANTS AND PRODUCTS STRAIGHT OFF THE TWO HALF-EQUATIONS

WORKEDEXAMPLE1—FECL3(AQ)+CU

Species present: Fe3+(aq), Cl−(aq), Cu(s)

HALF-EQUATION E° / V AVAILABLE?

Cu+ + e− ⇌ Cu +0.52 no Cu+ present

Cu2+ + 2e− ⇌ Cu +0.34 lowest available → Cu oxidised

Fe3+ + 3e− ⇌ Fe −0.04 would need Fe metal

Fe3+ + e− ⇌ Fe2+ +0.77 highest available → Fe3+ reduced

Cl2 + 2e− ⇌ 2Cl− +1.36 no Cl2 present

2 Fe3+ + Cu → Cu2+ + 2 Fe2+

STEPS 1–2: IDENTIFY SPECIES, THEN LIST EVERY RELEVANT ELECTRODE REACTION

WORKEDEXAMPLE2—FECL2+H2O2/H+

Species present: Fe2+, Cl−, H2O2, H+ reduction (highest) H2O2 + 2H+ + 2e− → 2H2O E° = +1.77 V oxidation (lowest available) 2Fe2+ → 2Fe3+ + 2e− E° = +0.77 V

2 Fe2+ + H2O2 + 2 H+ → 2 Fe3+ + 2 H2O

Fe2+ + 2e− ⇌ Fe (−0.44 V) is rejected: it would require Fe2+ to be reduced, but the higher-E° H2O2 takes the reduction role.

STEP 3: PICK HIGHEST AND LOWEST — AND CHECK THEY ARE ACTUALLY PRESENT

WORKEDEXAMPLE3—CRCL2+K2CR2O7/H+

Species present: Cr2+, Cl−, K+, Cr2O72−, H+ reduction Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O E° = +1.33 V oxidation 6 Cr2+ → 6 Cr3+ + 6e− E° = −0.41 V

6 Cr2+ + Cr2O72− + 14 H+ → 8 Cr3+ + 7 H2O

E°cell = 1.33 − (−0.41) = +1.74 V

Cl2/Cl− (+1.36 V) is rejected — no Cl2 present; K+/K (−2.92 V) is rejected — no K metal present.

REJECTING THE WRONG HALF-EQUATIONS — THE CROSSINGS- OUT MATTER

WORKEDEXAMPLE4—FECL3(AQ)+KI(AQ)

Species present: Fe3+, Cl−, K+, I− reduction 2 Fe3+ + 2e− → 2 Fe2+ E° = +0.77 V oxidation 2 I− → I2 + 2e− E° = +0.54 V

2 Fe3+ + 2 I− → 2 Fe2+ + I2

E°cell = 0.77 − 0.54 = +0.23 V

CHROMIUM(II) WITH DICHROMATE — SPECTATORS ELIMINATED ONE BY ONE

Chain Reactions with Excess Reagent

If one reagent is in excess, the product of step 1 can be attacked again. Work step by step, then add the steps and cancel.

IRON(III) CHLORIDE WITH POTASSIUM IODIDE, START TO FINISH

S N + E XC E SS I 2

Step 1 — from I2/I− (+0.54 V) and Sn2+/Sn (−0.14 V):

I2 + Sn → Sn2+ + 2 I− E°cell = 0.54 − (−0.14) = +0.68 V

Step 2 — Sn2+ now meets the remaining excess I2; from I2/I− (+0.54 V) and Sn4+/Sn2+ (+0.15 V):

Sn2+ + I2 → 2 I− + Sn4+

Step 1 + Step 2, cancelling Sn2+:

Sn + 2 I2 → 4 I− + Sn4+

TWO -STEP CHAIN, THEN THE STEPS ADDED AND CANCELLED

E XC E SS F E 3 + + V

Step 1 — Fe3+/Fe2+ (+0.77 V) vs V2+/V (−1.20 V):

2 Fe3+ + V → 2 Fe2+ + V2+

Step 2 — Fe3+/Fe2+ (+0.77 V) vs V3+/V2+ (−0.26 V):

Fe3+ + V2+ → Fe2+ + V3+

Step 3 — Fe3+/Fe2+ (+0.77 V) vs VO2+/V3+ (+0.34 V):

Fe3+ + V3+ + H2O → Fe2+ + VO2+ + 2 H+

Vanadium is oxidised in stages: V → V2+ → V3+ → VO2+, as long as Fe3+ remains in excess.

VANADIUM CLIMBING ONE OXIDATION STATE AT A TIME

Non-Standard Conditions & the Nernst Equation

Away from 1 mol dm−3, apply Le Chatelier's principle to the electrode equilibrium.

REASONING

For Fe3+ + e− ⇌ Fe2+, lowering [Fe3+] shifts the equilibrium left. More electrons are produced on the electrode, so the electrode potential decreases (becomes less than +0.77 V).

N E R N S T E Q UAT I O N

E = E° + 0.059 z lg [oxidised] [reduced]

z = number of electrons transferred in the half-equation

[oxidised] = species on the left of the half-equation

[reduced] = species on the right

[solid] = 1 — pure solids and the electrode metal do not appear

E X A M P L E — F E 3 + AT 0 . 0 0 1 M O L D M −3 , F E 2 + AT 1 M O L D M −3

E = 0.77 + (0.059 / 1) × lg(0.001 / 1)

E = 0.593 V

lower than E°, as Le Chatelier predicts

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E X A M P L E — C U 2 + AT 0 . 0 0 0 1 M O L D M −3

Cu2+ + 2e− ⇌ Cu, E° = +0.34 V, z = 2, [Cu] = 1 (solid).

E = 0.34 + (0.059 / 2) × lg(0.0001 / 1)

E = 0.22 V

Concentration decreased → equilibrium shifts left → potential becomes lower / more negative.

LE CHATELIER REASONING AND THE NERNST EQUATION, ANNOTATED

Homogeneous Catalysis

DEFINITION

A homogeneous catalyst is in the same phase as the reactants.

Example 1 — NO2 in the Chamber process

SO2 + NO2 → SO3 + NO

NO + ½ O2 → NO2

NO2 is regenerated — a homogeneous catalyst, since everything is gaseous.

THE COPPER EXAMPLE, WITH Z = 2

Example 2 — Fe2+/Fe3+ catalysing S2O82− + I−

MEMORISETHISONE

Uncatalysed: 2 I− + S2O82− → I2 + 2 SO42−, with E°cell = 2.01 − 0.54 = +1.47 V.

Yet no reaction occurs. Both ions are negatively charged and repel each other, so the activation energy is very high — despite the large, favourable E°cell.

Adding Fe2+/Fe3+ provides a two-step route in which each step involves oppositely charged ions:

2 I− + 2 Fe3+ → I2 + 2 Fe2+

S2O82− + 2 Fe2+ → 2 SO42− + 2 Fe3+

Fe2+/Fe3+ are regenerated — unchanged overall.

Ions are oppositely charged in both steps → attraction → much lower Ea → much faster.

Relevant potentials: I2/I− +0.54 V · Fe3+/Fe2+ +0.77 V · S2O82−/SO42− +2.01 V. The Fe couple sits between the other two — that is why it works.

HOMOGENEOUS CATALYST DEFINED, WITH THE NO 2 EXAMPLE

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WHY A BIG E° C E L L STILL GIVES NO REACTION — AND HOW FE 2 +/FE 3 + FIXES IT

Electrolysis

DEFINITION

The opposite of an electrochemical cell: electrical energy is converted into chemical energy — the decomposition of an electrolyte using electrical energy.

ELECTRODE WHAT HAPPENS WHICH SPECIES DIRECTION OF EQUATION

Anode (+) Oxidation — loses e− The one with the lower E° Written backwards

Cathode (−) Reduction — gains e− The one with the higher E° Written forwards

Dilute NaCl(aq)

Species present: Na+, Cl−, H2O. Note that water ionises very little, so H2O itself is the species discharged.

anode 2 H2O → 4 H+ + O2 + 4e−

ELECTROLYSIS DEFINED AS THE REVERSE OF A CELL cathode 2 H2O + 2e− → H2 + 2 OH−

CuSO4(aq)

Species present: Cu2+, SO42−, H2O.

anode 2 H2O → 4 H+ + O2 + 4e− E° = +1.23 V cathode Cu2+ + 2e− → Cu E° = +0.34 V

ANODE AND CATHODE L ABELLED BY E° — DILUTE NACL

C L A SS I C E X A M Q U E S T I O N — W H Y I S C L − D I S C H A R G E D F R O M C O N C E N T R AT E D N A C L ?

At the anode the candidates are Cl− and H2O:

Cl2 + 2e− ⇌ 2 Cl− E° = +1.36 V standard

O2 + 4H+ + 4e− ⇌ 2 H2O E° = +1.23 V

Under standard conditions O2/H2O has the lower E°, so water should be oxidised. But when Cl− is concentrated, the Cl2/Cl− equilibrium shifts to the left, so its electrode potential decreases (to roughly +1.20 V) and becomes the lowest. Hence Cl− is discharged and chlorine gas is produced.

COPPER(II) SULFATE ELECTROLYSIS

The Faraday Constant

QUANTITY VALUE

Charge on one electron 1.6 × 10−19 C

Avogadro constant, L 6.02 × 1023 mol−1

Faraday constant, F = L × e 96 500 C mol−1

T H E T W O E Q UAT I O N S

1 mol e− ≡ 96 500 C Q=I×t

(coulombs = amperes × seconds)

THE FULL ARGUMENT, WRITTEN OUT — CONCENTRATION REORDERS THE POTENTIALS

W O R K E D E X A M P L E — C U S O 4 E L E C T R O LY S E D AT 2 A F O R 3 0 M I N U T E S . F I N D T H E

M A SS O F C U D E P O S I T E D .

DERIVING F FROM THE AVOGADRO CONSTANT AND THE ELECTRONIC CHARGE

Q = I × t = 2 × (30 × 60) = 3600 C mol e− = 3600 / 96 500 = 0.0373 mol

Cu2+ + 2e− → Cu, so mol Cu = 0.0373 / 2 = 0.0187 mol m = n × Mr = 0.0187 × 63.5 = 1.184 g

METHODINONELINE

Q = I t → ÷ 96 500 for mol e− → ÷ (electrons in the half-equation) for mol product → × Mr for mass

(or ÷ 24 000 cm3 / 24 dm3 for gas volume at r.t.p.).

THE FULL WORKED ANSWER, STEP BY STEP

Exam Checklist

Quote standard conditions in full: 298 K, 1 atm, 1 mol dm−3.

Define E° as a potential difference against a standard hydrogen electrode — not just "the voltage".

Higher E° → reduced (gains e−); lower E° → oxidised (loses e−). Electrons flow towards higher E°.

E°cell = E°(reduction) − E°(oxidation). Positive → feasible.

Feasible ≠ fast. A high E°cell can still give no reaction if Ea is large (S2O82− + I−).

Always check that the species you assign are actually present in the mixture.

For non-standard concentrations, argue with Le Chatelier first, then use Nernst if numbers are required.

In electrolysis, remember the anode reaction is written backwards (as oxidation).

Concentration changes can reorder electrode potentials — that is the whole point of the concentrated-

NaCl question.

Faraday problems: never forget to divide by the number of electrons in the half-equation.

Fahad H. Ahmad

C H E M I S T RY L E A D · M E G A L E C T U R E

Need a walkthrough of this topic, past-paper drilling, or a full A2 Chemistry crash course? Book a free demo class —

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