Electrochemistry — Cells, Nernst Equation & Conductance

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Electrochemistry links two things you already know: redox reactions and thermodynamics. A cell is a redox reaction with the electrons sent through a wire. Once you see that, every formula in this chapter connects back to ΔG.

What you will be able to do

• Write cell notation, pick the anode and cathode, and calculate E°cell from the series
• Use the Nernst equation for any concentration, including concentration cells and pH
• Move between E°, ΔG° and K, and explain why E° values can’t simply be added
• Calculate κ, Λm, α and Ka from conductivity data using Kohlrausch’s law
• Predict electrolysis products and solve every type of Faraday’s-law numerical

The big picture

Two kinds of cell, one redox reaction. Everything else in the chapter is bookkeeping for the electrons.

In one line: A spontaneous redox reaction pushes electrons through a wire (galvanic cell); a battery can push them the other way (electrolysis). E°, ΔG° and K are three views of the same driving force.

Think of it like this: A galvanic cell is a waterfall: electrons fall downhill from the metal that gives them up easily (zinc) to the ion that grabs them (Cu²⁺), and you put a turbine in the wire. Electrolysis is a pump: you spend energy to push them uphill.

	Galvanic (voltaic) cell	Electrolytic cell
Energy change	Chemical → electrical	Electrical → chemical
Reaction	Spontaneous, ΔG < 0	Non-spontaneous, driven by an external source
Anode (oxidation)	Negative	Positive
Cathode (reduction)	Positive	Negative
Examples	Daniell cell, dry cell, fuel cell	Electroplating, extraction of Na and Al, refining Cu

Key idea: In both cells, oxidation is at the anode and reduction at the cathode. Only the signs change. Remember “An Ox, Red Cat”.

Cells and cell notation

How to write a cell, and what the salt bridge is really doing.

Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)

Anode on the left, cathode on the right. | = phase boundary, ‖ = salt bridge

• Anode reaction: Zn → Zn²⁺ + 2e⁻ (oxidation)
• Cathode reaction: Cu²⁺ + 2e⁻ → Cu (reduction)
• Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu
• Salt bridge (KCl or KNO₃ in agar): completes the circuit and keeps both solutions electrically neutral. Without it, the zinc side builds up positive charge and the current stops within seconds.

Exam trap: Students write the salt bridge as a source of electrons. It carries ions, not electrons. Electrons travel only through the external wire.

Electrode potentials and the series

One number per half-cell tells you who gives electrons to whom.

The standard hydrogen electrode (SHE) is the zero of the scale: 2H⁺(aq, 1 M) + 2e⁻ → H₂(g, 1 bar), E° = 0.00 V at 298 K. Every other standard reduction potential is measured against it.

Half-reaction (reduction)	E° (V)	What it tells you
F₂ + 2e⁻ → 2F⁻	+2.87	Strongest oxidising agent
Cl₂ + 2e⁻ → 2Cl⁻	+1.36
O₂ + 4H⁺ + 4e⁻ → 2H₂O	+1.23
Ag⁺ + e⁻ → Ag	+0.80	Silver is a poor reducing agent
Cu²⁺ + 2e⁻ → Cu	+0.34	Cu does not displace H₂ from acids
2H⁺ + 2e⁻ → H₂	0.00	Reference
Fe²⁺ + 2e⁻ → Fe	−0.44	Fe displaces H₂ from dilute acids
Zn²⁺ + 2e⁻ → Zn	−0.76
Na⁺ + e⁻ → Na	−2.71
Li⁺ + e⁻ → Li	−3.05	Strongest reducing agent

E°cell = E°(cathode) − E°(anode)

Use reduction potentials for both. Positive E°cell → spontaneous as written.

Four things the series predicts

1. Which electrode is the anode: the one with the lower (more negative) E°.
2. Displacement: a metal displaces another from solution if it is lower in the series. Zn displaces Cu from CuSO₄; Cu does not displace Zn.
3. Reaction with acids: metals with E° < 0 release H₂ from dilute acids (Zn, Fe); metals with E° > 0 don’t (Cu, Ag).
4. Storage: CuSO₄ solution can’t be kept in a zinc pot (zinc would dissolve), but can be kept in a silver one.

Solved example 01 — Choosing the anode

A cell is made from Ag⁺/Ag (E° = +0.80 V) and Cu²⁺/Cu (E° = +0.34 V). Identify the anode and find E°cell.

1. Lower E° → oxidised → anode: Cu.
2. Cathode: Ag⁺ + e⁻ → Ag.
3. E°cell = 0.80 − 0.34 = 0.46 V.

Answer: Cu is the anode; E°cell = 0.46 V; cell reaction Cu + 2Ag⁺ → Cu²⁺ + 2Ag.

Check yourself

• C1. Can you store AgNO₃ solution in a copper vessel?
• C2. Which is the stronger reducing agent: Zn or Fe?
• C3. Will dilute HCl react with silver?

Answers at the end.

The Nernst equation

E° is for 1 M solutions. Real cells aren’t standard; the Nernst equation corrects for that.

E = E° − (0.059 / n) log Q (at 298 K)

n = electrons in the balanced cell reaction. Pure solids and liquids are left out of Q.

For the Daniell cell, Zn + Cu²⁺ → Zn²⁺ + Cu, so Q = [Zn²⁺] / [Cu²⁺].

Key idea: Read the equation as a see-saw: more reactant ions (smaller Q) → higher E; more product ions (larger Q) → lower E. At equilibrium the cell is flat: E = 0.

Three special cases examiners love

• Concentration cell: the same electrode in two concentrations, so E° = 0 and E = (0.059/n) log(c(cathode)/c(anode)). The more concentrated side is the cathode.
• Hydrogen electrode at other pH: E = −0.059 × pH (with H₂ at 1 bar). At pH 10, E = −0.59 V.
• Metal ion electrode: E(M²⁺/M) = E° − (0.059/2) log(1/[M²⁺]). Diluting the ions makes the potential more negative.

Solved example 02 — Nernst with real concentrations

Find E for Mg | Mg²⁺(0.001 M) ‖ Cu²⁺(0.0001 M) | Cu. E°cell = 2.71 V.

1. Mg + Cu²⁺ → Mg²⁺ + Cu, n = 2.
2. Q = [Mg²⁺]/[Cu²⁺] = 0.001/0.0001 = 10.
3. E = 2.71 − (0.059/2) × log 10 = 2.71 − 0.0295.

Answer: E ≈ 2.68 V.

Solved example 03 — A concentration cell

Find E for Cu | Cu²⁺(0.01 M) ‖ Cu²⁺(0.1 M) | Cu.

1. E° = 0 (same electrode both sides), n = 2.
2. E = (0.059/2) log(0.1/0.01) = 0.0295 × 1.

Answer: E = 0.0295 V; the 0.1 M side is the cathode.

Check yourself

• C4. What happens to E of a Daniell cell if [Zn²⁺] is increased?
• C5. A hydrogen electrode dips into a solution of pH 3. What is its potential?

Answers at the end.

The E°–ΔG°–K triangle

Three quantities, one driving force. Know any one and you have the other two.

ΔG° = −nFE°
E°cell = (0.059 / n) log K (at 298 K)
ΔG° = −2.303 RT log K

F = 96,487 C mol⁻¹ (≈ 96,500)

E°cell	ΔG°	K	Meaning
Positive	Negative	> 1	Products favoured; spontaneous
Zero	Zero	1	At equilibrium
Negative	Positive	< 1	Reactants favoured

Exam trap: E° is intensive; ΔG° is extensive. Doubling a half-reaction doubles ΔG° but leaves E° unchanged. So you can add ΔG° values, but never add E° values of half-reactions directly.

Solved example 04 — K from E°

For the Daniell cell, E° = 1.10 V. Find K and ΔG°.

1. log K = nE°/0.059 = 2 × 1.10/0.059 = 37.29.
2. K = antilog(37.29) ≈ 1.9 × 10³⁷.
3. ΔG° = −2 × 96,487 × 1.10 = −212,271 J mol⁻¹.

Answer: K ≈ 1.9 × 10³⁷; ΔG° ≈ −212.3 kJ mol⁻¹. The reaction goes essentially to completion.

Conductance

How well a solution carries current, and how to use that to measure dissociation.

Quantity	Symbol and unit	What it means
Resistance	R (Ω)	Opposition to current
Conductance	G = 1/R (S)	Ease of current flow
Cell constant	G* = l/A (cm⁻¹)	Geometry of the conductivity cell
Conductivity	κ = G* / R (S cm⁻¹)	Conductance of a 1 cm cube of solution
Molar conductivity	Λm = κ × 1000 / M (S cm² mol⁻¹)	Conductance of 1 mol of electrolyte

What happens on dilution

• κ decreases: fewer ions per cm³.
• Λm increases: ions move more freely, and weak electrolytes dissociate more.
• Strong electrolyte: Λm = Λ°m − A√c. A plot of Λm against √c is a straight line; extrapolate to c = 0 to get Λ°m.
• Weak electrolyte: Λm rises steeply near zero concentration, so it can’t be extrapolated. Use Kohlrausch’s law.

Λ°m = ν₊ λ°₊ + ν₋ λ°₋

Kohlrausch: at infinite dilution, each ion contributes independently

Ion	λ° (S cm² mol⁻¹)
H⁺	349.6
OH⁻	199.1
Na⁺	50.1
Cl⁻	76.3
CH₃COO⁻	40.9

Why are H⁺ and OH⁻ so fast? They don’t travel themselves; protons hop from one water molecule to the next along hydrogen bonds (the Grotthuss mechanism).

Uses of Kohlrausch’s law

1. Λ°m of a weak electrolyte: Λ°(CH₃COOH) = Λ°(HCl) + Λ°(CH₃COONa) − Λ°(NaCl).
2. Degree of dissociation: α = Λm / Λ°m.
3. Dissociation constant: Ka = cα² / (1 − α).

Solved example 05 — Cell constant and Λm

A cell filled with 0.1 M KCl (κ = 0.0129 S cm⁻¹) has R = 100 Ω. Filled with 0.02 M of another electrolyte, R = 520 Ω. Find Λm of the second solution.

1. G* = κR = 0.0129 × 100 = 1.29 cm⁻¹.
2. κ(new) = G*/R = 1.29/520 = 2.48 × 10⁻³ S cm⁻¹.
3. Λm = κ × 1000/M = 2.48 × 10⁻³ × 1000/0.02.

Answer: Λm ≈ 124 S cm² mol⁻¹.

Solved example 06 — α and Ka of acetic acid

0.001028 M acetic acid has κ = 4.95 × 10⁻⁵ S cm⁻¹. Λ°m(CH₃COOH) = 390.5 S cm² mol⁻¹.

1. Λm = 4.95 × 10⁻⁵ × 1000/0.001028 = 48.15 S cm² mol⁻¹.
2. α = 48.15/390.5 = 0.1233.
3. Ka = cα²/(1 − α) = 0.001028 × 0.1233²/0.8767.

Answer: α ≈ 0.123; Ka ≈ 1.78 × 10⁻⁵.

Check yourself

• C6. Why does κ fall but Λm rise on dilution?
• C7. Find Λ°m of CH₃COOH from the ion table above.

Answers at the end.

Electrolysis and Faraday’s laws

Pushing electrons uphill, and counting exactly how much you get for your charge.

m = (M / nF) × I × t

First law: mass ∝ charge. Second law: for the same charge, masses ∝ M/n.

1 F = 96,487 C ≈ 96,500 C = the charge on 1 mol of electrons. Charge Q = I × t (amperes × seconds).

Predicting the products

Electrolyte	Cathode	Anode	Why
Molten NaCl	Na	Cl₂	Only Na⁺ and Cl⁻ present
Aqueous NaCl (concentrated)	H₂	Cl₂	Water (−0.83 V) is far easier to reduce than Na⁺ (−2.71 V); O₂ has a high overpotential, so Cl₂ forms
Dilute H₂SO₄	H₂	O₂	Water is oxidised; SO₄²⁻ is very hard to oxidise
Concentrated H₂SO₄	H₂	S₂O₈²⁻	At high concentration, HSO₄⁻ is oxidised to peroxodisulphate
Aqueous CuSO₄, Pt electrodes	Cu	O₂	Cu²⁺ is easier to reduce than water
Aqueous CuSO₄, Cu electrodes	Cu deposits	Cu dissolves	Basis of copper refining and electroplating

Key idea: At the cathode, the species with the highest reduction potential is reduced. At the anode, the species easiest to oxidise is oxidised, unless overpotential intervenes (O₂ from water usually needs extra voltage).

Solved example 07 — Mass deposited

2.0 A flows through CuSO₄ solution for 965 s. Find the mass of copper deposited (Cu = 63.5).

1. Q = 2.0 × 965 = 1930 C = 1930/96,500 = 0.020 mol e⁻.
2. Cu²⁺ + 2e⁻ → Cu, so mol Cu = 0.010.
3. Mass = 0.010 × 63.5.

Answer: 0.635 g of copper.

Solved example 08 — Time required

How long does a 2.0 A current take to deposit 1.27 g of copper?

1. mol Cu = 1.27/63.5 = 0.020; mol e⁻ = 0.040.
2. Q = 0.040 × 96,500 = 3860 C.
3. t = Q/I = 3860/2.0 = 1930 s.

Answer: 1930 s ≈ 32.2 min.

Solved example 09 — Volume of gas and the second law

965 C is passed through dilute H₂SO₄. What volume of O₂ (at STP) forms at the anode? If the same charge passes through AgNO₃, what mass of Ag deposits?

1. 965 C = 0.010 mol e⁻.
2. 2H₂O → O₂ + 4H⁺ + 4e⁻: mol O₂ = 0.010/4 = 0.0025 → 0.0025 × 22,400 mL.
3. Ag⁺ + e⁻ → Ag: mol Ag = 0.010 → 0.010 × 108 g.

Answer: 56 mL of O₂; 1.08 g of Ag.

Batteries, fuel cells and corrosion

The same chemistry in your torch, your car and on a rusting gate.

Device	Anode	Cathode	Notes
Dry (Leclanché) cell	Zn → Zn²⁺ + 2e⁻	MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃	About 1.5 V; not rechargeable
Mercury cell	Zn(Hg)	HgO → Hg	Steady 1.35 V; hearing aids, watches
Lead storage battery	Pb	PbO₂	Discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O; rechargeable
H₂–O₂ fuel cell	2H₂ + 4OH⁻ → 4H₂O + 4e⁻	O₂ + 2H₂O + 4e⁻ → 4OH⁻	Product is water; used in spacecraft; about 70% efficient

Corrosion (rusting) is a tiny galvanic cell on the iron surface. At one spot (the anode) Fe → Fe²⁺ + 2e⁻; at another (the cathode) O₂ + 4H⁺ + 4e⁻ → 2H₂O. Fe²⁺ is later oxidised to rust, Fe₂O₃·xH₂O. Salt water speeds it up by carrying ions.

• Barrier protection: paint, oil, plastic coating.
• Sacrificial protection: galvanising with zinc, or attaching Mg blocks to ships and pipelines. The more reactive metal is oxidised instead of iron, even if the coating is scratched.

Exam trap gallery

The mistakes that cost the most marks in this chapter, side by side with the fix.

#	What students write	What is actually correct	Why
1	Doubling a half-reaction doubles its E°.	E° stays the same; only ΔG° doubles.	E° is intensive (per electron); ΔG° is extensive.
2	E°(Fe³⁺/Fe) = E°(Fe³⁺/Fe²⁺) + E°(Fe²⁺/Fe) = 0.77 + (−0.44) = 0.33 V.	Add ΔG°, then divide: E° = (1 × 0.77 + 2 × (−0.44))/3 = −0.037 V.	E° values of half-reactions can’t be added; their ΔG° values can.
3	In a galvanic cell, the anode is positive.	In a galvanic cell the anode is negative; in an electrolytic cell it is positive.	Oxidation is always at the anode; only the sign changes.
4	E°cell = E°(anode) − E°(cathode).	E°cell = E°(cathode) − E°(anode), both as reduction potentials.	Positive E°cell must mean spontaneous.
5	On dilution, both κ and Λm increase.	κ decreases; Λm increases.	Fewer ions per cm³, but each mole of electrolyte conducts better.
6	Aqueous NaCl gives Na at the cathode.	It gives H₂; only molten NaCl gives Na.	Water is reduced far more easily than Na⁺.
7	Λm = κ/M.	Λm = κ × 1000/M when κ is in S cm⁻¹ and M in mol L⁻¹.	The 1000 converts litres to cm³.
8	The salt bridge supplies electrons to the cathode.	The salt bridge carries ions to keep each half-cell neutral.	Electrons flow only through the external wire.

Memory hook: “An Ox, Red Cat” — Anode Oxidation, Reduction Cathode. · “LEO says GER” — Lose Electrons Oxidation, Gain Electrons Reduction. · “E° stays, ΔG° pays” — only ΔG° scales with the equation. · “κ falls, Λ climbs” — what dilution does.

Practice MCQs

Levels: Easy checks recall, Moderate needs one reasoning step, Tough combines two ideas. Full explanations follow.

Q01. In a galvanic cell, electrons flow: (Easy)

• (a) from cathode to anode through the salt bridge
• (b) from anode to cathode through the external wire
• (c) from cathode to anode through the external wire
• (d) in both directions

Q02. The function of a salt bridge is to: (Easy)

• (a) supply electrons
• (b) maintain electrical neutrality in each half-cell
• (c) increase the cell voltage
• (d) act as a catalyst

Q03. Which metal can displace copper from CuSO₄ solution? (Easy)

• (a) Ag
• (b) Au
• (c) Zn
• (d) Pt

Q04. E° for Cu²⁺/Cu is +0.34 V and for Ag⁺/Ag is +0.80 V. E°cell for the cell made from them is: (Easy)

• (a) 1.14 V
• (b) 0.46 V
• (c) −0.46 V
• (d) 0.23 V

Q05. The unit of molar conductivity is: (Easy)

• (a) S cm⁻¹
• (b) S cm² mol⁻¹
• (c) S mol cm⁻²
• (d) Ω cm

Q06. On dilution of an aqueous KCl solution: (Moderate)

• (a) κ and Λm both increase
• (b) κ increases, Λm decreases
• (c) κ decreases, Λm increases
• (d) both decrease

Q07. The potential of a hydrogen electrode (H₂ at 1 bar) in a solution of pH 4 at 298 K is: (Moderate)

• (a) +0.236 V
• (b) −0.236 V
• (c) −0.059 V
• (d) 0.00 V

Q08. For Zn | Zn²⁺(0.1 M) ‖ Ag⁺(0.01 M) | Ag, E°cell = 1.56 V. Ecell at 298 K is about: (Moderate)

• (a) 1.56 V
• (b) 1.62 V
• (c) 1.47 V
• (d) 1.50 V

Q09. Electrolysis of aqueous NaCl gives at the cathode: (Moderate)

• (a) Na
• (b) Cl₂
• (c) H₂
• (d) O₂

Q10. The charge needed to deposit 1 mol of Al from Al³⁺ is: (Moderate)

• (a) 1 F
• (b) 2 F
• (c) 3 F
• (d) 6 F

Q11. When a lead storage battery discharges: (Moderate)

• (a) PbSO₄ is formed at both electrodes
• (b) H₂SO₄ concentration increases
• (c) PbO₂ is formed
• (d) Pb is deposited

Q12. Which protects iron even when the coating is scratched? (Moderate)

• (a) Tin plating
• (b) Painting
• (c) Zinc coating
• (d) Oiling

Q13. E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Fe²⁺/Fe) = −0.44 V. E°(Fe³⁺/Fe) is: (Tough)

• (a) +0.33 V
• (b) −0.037 V
• (c) +1.21 V
• (d) −0.11 V

Q14. For the cell reaction Cu + 2Ag⁺ → Cu²⁺ + 2Ag, E° = 0.46 V. log K at 298 K is about: (Tough)

• (a) 7.8
• (b) 15.6
• (c) 31.2
• (d) 3.9

Q15. A current is passed through AgNO₃ and CuSO₄ solutions in series. 1.08 g Ag is deposited. The mass of Cu deposited is: (Tough)

• (a) 0.635 g
• (b) 0.3175 g
• (c) 1.27 g
• (d) 0.108 g

Answer key with explanations

1. (b) from anode to cathode through the external wire — Electrons leave the anode (oxidation) and travel through the wire to the cathode.
2. (b) maintain electrical neutrality in each half-cell — It carries ions, not electrons.
3. (c) Zn — Zn (E° = −0.76 V) is lower in the series than Cu.
4. (b) 0.46 V — 0.80 − 0.34 = 0.46 V; Cu is the anode.
5. (b) S cm² mol⁻¹ — Λm = κ × 1000/M gives S cm² mol⁻¹.
6. (c) κ decreases, Λm increases — Fewer ions per cm³, but each mole conducts better.
7. (b) −0.236 V — E = −0.059 × pH = −0.236 V.
8. (c) 1.47 V — Zn + 2Ag⁺ → Zn²⁺ + 2Ag, n = 2, Q = 0.1/(0.01)² = 1000; E = 1.56 − 0.0295 × 3 = 1.47 V.
9. (c) H₂ — Water is reduced far more easily than Na⁺.
10. (c) 3 F — Al³⁺ + 3e⁻ → Al.
11. (a) PbSO₄ is formed at both electrodes — Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O; the acid is used up.
12. (c) Zinc coating — Zinc is more reactive and is oxidised first (sacrificial). A scratched tin coating speeds rusting.
13. (b) −0.037 V — Add ΔG°: −3FE = −F(0.77) − 2F(−0.44) → E = −0.037 V.
14. (b) 15.6 — log K = nE°/0.059 = 2 × 0.46/0.059 = 15.6.
15. (b) 0.3175 g — Same charge: 0.010 mol e⁻ → 0.005 mol Cu = 0.3175 g.

Advanced problems

TGT/PGT and JEE level: each combines two ideas from this chapter.

Solved example 10 — Non-additive E°

Given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Fe²⁺/Fe) = −0.44 V, find E°(Fe³⁺/Fe).

1. Write ΔG° for each: Fe³⁺ + e⁻ → Fe²⁺, ΔG°₁ = −1 × F × 0.77.
2. Fe²⁺ + 2e⁻ → Fe, ΔG°₂ = −2 × F × (−0.44) = +0.88F.
3. Fe³⁺ + 3e⁻ → Fe: ΔG°₃ = ΔG°₁ + ΔG°₂ = −0.77F + 0.88F = +0.11F.
4. E°₃ = −ΔG°₃/3F = −0.11/3.

Answer: E°(Fe³⁺/Fe) ≈ −0.037 V.

Solved example 11 — Nernst with unequal coefficients

Find Ecell for Zn | Zn²⁺(0.1 M) ‖ Ag⁺(0.01 M) | Ag. E°(Ag⁺/Ag) = +0.80 V, E°(Zn²⁺/Zn) = −0.76 V.

1. E°cell = 0.80 − (−0.76) = 1.56 V.
2. Zn + 2Ag⁺ → Zn²⁺ + 2Ag, so n = 2 and Q = [Zn²⁺]/[Ag⁺]² = 0.1/(0.01)² = 1000.
3. E = 1.56 − (0.059/2) × log 1000 = 1.56 − 0.0885.

Answer: Ecell ≈ 1.47 V. (The trap: forgetting to square [Ag⁺].)

Check-yourself answers

• C1. No. Cu (E° = +0.34 V) is lower than Ag (+0.80 V), so copper would reduce Ag⁺ and dissolve.
• C2. Zn (E° = −0.76 V vs −0.44 V for Fe).
• C3. No. Silver has E° > 0, so it can’t reduce H⁺.
• C4. E decreases (Q = [Zn²⁺]/[Cu²⁺] gets larger).
• C5. E = −0.059 × 3 = −0.177 V.
• C6. κ counts ions per cm³ (fewer on dilution); Λm is per mole, and each mole conducts better when ions are far apart.
• C7. Λ° = 349.6 + 40.9 = 390.5 S cm² mol⁻¹.

One-page revision

Read this page the night before the exam. Every line points back to a section in these notes.

Cells

• Galvanic: chemical → electrical, ΔG < 0, anode −
• Electrolytic: electrical → chemical, anode +
• Oxidation always at the anode (An Ox, Red Cat)
• Salt bridge carries ions, not electrons

Potentials

• E°cell = E°(cathode) − E°(anode)
• Lower E° = anode = better reducing agent
• E < 0 metals release H₂ from dilute acid
• E° intensive; add ΔG°, never E°

Nernst and the triangle

• E = E° − (0.059/n) log Q
• Concentration cell: E = (0.059/n) log(c₂/c₁)
• H₂ electrode: E = −0.059 pH
• ΔG° = −nFE°; log K = nE°/0.059

Conductance

• κ = G*/R; Λm = κ × 1000/M
• Dilution: κ falls, Λm rises
• Strong: Λm = Λ°m − A√c
• α = Λm/Λ°m; Ka = cα²/(1 − α)

Electrolysis

• m = (M/nF) × I × t; 1 F = 96,500 C
• Aq. NaCl → H₂ + Cl₂; molten NaCl → Na + Cl₂
• Dilute H₂SO₄ → H₂ + O₂
• Same charge: masses ∝ M/n

Devices and corrosion

• Dry cell 1.5 V; mercury cell 1.35 V steady
• Lead battery: PbSO₄ on both plates when discharged
• Fuel cell product: water
• Zinc protects iron sacrificially