Electrochemistry — Cells, Nernst Equation and Electrolysis

Electrochemistry is where redox chemistry, thermodynamics and kinetics meet, and it is unusually rewarding because a single sign convention unlocks most of it. Get one thing straight at the outset: oxidation always happens at the anode and reduction at the cathode, in every cell of every type. What changes between a galvanic cell and an electrolytic cell is only the polarity — the anode is negative in a galvanic cell and positive in an electrolytic one. Candidates who memorise "anode is negative" get half the questions in this chapter wrong; candidates who memorise "anode is oxidation" get none of them wrong for that reason.

Galvanic Cells and Electrode Potential

A galvanic cell converts the free energy of a spontaneous redox reaction into electrical work. By convention the cell is written anode ‖ cathode, left to right, as in Zn | Zn²⁺ (1 M) ‖ Cu²⁺ (1 M) | Cu. The cell potential is E°_cell = E°_cathode − E°_anode, using reduction potentials for both, and a positive E°_cell means the reaction as written is spontaneous.

Galvanic versus electrolytic — what changes and what does not
FeatureGalvanic cellElectrolytic cell
Anode reactionOxidationOxidation — the same
Anode polarityNegativePositive
Energy flowChemical → electricalElectrical → chemical
ΔG of the cell reactionNegative — spontaneousPositive — must be driven
ExampleDaniell cell, dry cell, lead storage battery on dischargeElectroplating, Hall–Héroult process, recharging a battery
🎯 Why the standard hydrogen electrode is set at exactly zero
A single electrode potential can never be measured, because any measurement requires a complete circuit and therefore a second electrode. What is measurable is only the difference between two electrodes. So chemists chose one electrode — 1 M H⁺, 1 bar H₂, platinum, 298 K — and assigned it 0.00 V by definition, exactly as sea level is assigned zero altitude. Every tabulated E° value is therefore a potential relative to hydrogen, not an absolute quantity. This is why the sign of E° tells you how a species behaves compared with hydrogen: a positive E° means it is reduced more readily than H⁺, which is precisely why copper (+0.34 V) does not dissolve in HCl while zinc (−0.76 V) does.
⚠️ E° is an intensive property — never multiply it when you scale the equation
Doubling a cell reaction doubles ΔG and doubles n, but leaves E° completely unchanged, because E° is energy per unit charge and both numerator and denominator scale together. So for Zn + Cu²⁺ → Zn²⁺ + Cu with E° = 1.10 V, the reaction 2Zn + 2Cu²⁺ → 2Zn²⁺ + 2Cu still has E° = 1.10 V, not 2.20 V. The option 2.20 V is offered in almost every question of this type. Reversing the reaction, by contrast, does flip the sign of E°. The same logic explains why connecting two identical cells in series doubles the voltage while a single bigger cell does not — that is circuitry, not thermodynamics.

The bridge to thermodynamics is ΔG° = −nFE°_cell, with F = 96 500 C/mol. Combining it with ΔG° = −RT ln K gives log K = nE°_cell/0.0591 at 298 K, so a cell potential of just 0.3 V with n = 2 corresponds to an equilibrium constant of about 10¹⁰ — electrochemical measurements are extraordinarily sensitive probes of equilibrium.

The Nernst Equation and Concentration Cells

Away from standard conditions the potential shifts, and the Nernst equation quantifies it: E = E° − (0.0591/n) log Q at 298 K, where Q is the reaction quotient built from products over reactants exactly as for equilibrium. Raising the product-ion concentration lowers E; raising the reactant-ion concentration raises it. Solids and pure liquids are omitted from Q, and for a gas its partial pressure is used.

🧠 At equilibrium E = 0, and that is when a battery is flat
Set E = 0 in the Nernst equation and you get E° = (0.0591/n) log K — the same relation as before, now with a physical meaning. As a cell discharges, product ions accumulate and reactant ions deplete, so Q climbs towards K and E falls steadily towards zero. A dead battery is a cell that has reached equilibrium, not one that has run out of substance. This also gives you a free result: for a concentration cell, where both electrodes are the same metal, E° = 0 and the entire potential comes from the log term, E = −(0.0591/n) log([dilute]/[concentrated]). Such a cell drives ions until the two concentrations equalise and then stops.

Conductance completes the physical picture. Conductivity κ is conductance per unit dimension and always falls on dilution, because there are fewer ions in a given volume. Molar conductivity Λ_m = κ × 1000/M always rises on dilution, because it counts the conductance per mole of electrolyte and dilution frees the ions from interionic drag. For a strong electrolyte the rise is small and linear, following Debye–Hückel–Onsager, and Λ°_m is found by extrapolating to zero concentration. For a weak electrolyte the rise is dramatic and steeply curved near zero, so extrapolation is impossible and Λ°_m must instead be assembled from ionic contributions using Kohlrausch's law of independent migration.

ℹ️ Kohlrausch's law lets you get Λ° for a weak acid from three strong salts
Because each ion contributes independently, Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl). Every quantity on the right belongs to a strong electrolyte and so is obtainable by extrapolation; the sodium and chloride contributions cancel, leaving exactly the acid. Once Λ°_m is known, the degree of dissociation is α = Λ_m/Λ°_m and then K_a = Cα²/(1 − α) — which is how dissociation constants of weak acids were first measured, long before pH meters existed.

Electrolysis, Faraday's Laws and Corrosion

Faraday's first law says the mass deposited is proportional to the charge passed: m = (Q/F) × (M/n) = (It/96500) × (M/n), where M/n is the equivalent mass. The second law says that for the same charge, the masses of different substances deposited stand in the ratio of their equivalent masses — which is why one faraday deposits 108 g of silver (n = 1) but only 32 g of copper (n = 2) and 9 g of aluminium (n = 3).

⚠️ Electrolysis of brine gives hydrogen and chlorine, not sodium
In an aqueous solution, water itself competes at both electrodes, and the species with the more favourable potential wins. At the cathode, reducing water to H₂ is far easier than reducing Na⁺ (−2.71 V), so hydrogen is evolved and sodium is never obtained from aqueous NaCl — sodium metal requires molten NaCl, which is exactly what the Downs cell uses. At the anode, oxidising Cl⁻ to Cl₂ competes with oxidising water to O₂; although water is thermodynamically favoured, chlorine wins in concentrated brine because of its much lower overpotential, a kinetic effect. So a question about aqueous versus molten electrolysis is asking whether you remembered that water is a reactant too.
The batteries the paper names
CellChemistryPoint examined
Dry (Leclanché) cellZn anode, MnO₂/C cathode, NH₄Cl pasteNot rechargeable; voltage falls as Zn is consumed
Mercury cellZn–Hg amalgam anode, HgO cathodeConstant 1.35 V, since no ion appears in the cell reaction
Lead storage batteryPb anode, PbO₂ cathode, 38 % H₂SO₄Rechargeable; acid is consumed so density drops on discharge
H₂–O₂ fuel cellPorous carbon electrodes in KOH, Pt catalystContinuous supply, efficiency about 70 %, product is water
🎯 Why zinc protects iron even when the coating is scratched
Rusting is an electrochemical process: iron oxidises at an anodic patch, oxygen is reduced at a cathodic patch, and water completes the circuit — which is why iron does not rust in dry air or in oxygen-free water, but rusts fastest in salty water, where the electrolyte is best. Painting works only while it is intact, because it simply blocks contact. Galvanising works differently and better: zinc has a more negative E° (−0.76 V) than iron (−0.44 V), so at any exposed scratch it is zinc, not iron, that becomes the anode and dissolves. The zinc is sacrificed to keep the iron cathodic. That is why a scratched galvanised sheet still does not rust, and it is the same principle as bolting magnesium blocks to a ship's hull or a buried pipeline.

Key takeaways

  • Oxidation is at the anode in every cell; only the polarity flips — negative in galvanic, positive in electrolytic.
  • E°_cell = E°_cathode − E°_anode, and a positive value means the reaction as written is spontaneous.
  • E° is intensive: scaling the equation leaves it unchanged, though reversing the equation flips its sign.
  • ΔG° = −nFE°_cell with F = 96 500 C/mol, and log K = nE°_cell/0.0591 at 298 K.
  • Nernst: E = E° − (0.0591/n) log Q. A flat battery is a cell that has reached equilibrium, where E = 0.
  • Conductivity κ falls on dilution while molar conductivity Λ_m rises; Kohlrausch gives Λ° for weak electrolytes.
  • Faraday: m = (It/96500)(M/n), so one faraday gives 108 g Ag but 32 g Cu and 9 g Al.
  • Aqueous NaCl electrolysis gives H₂ and Cl₂; sodium metal needs molten NaCl. Zinc protects iron by sacrificing itself.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In an electrolytic cell, the anode is:

    1. Negative, and reduction occurs there
    2. Positive, and oxidation occurs there
    3. Negative, and oxidation occurs there
    4. Positive, and reduction occurs there
    Show answer

    Answer: B — Positive, and oxidation occurs there

    Oxidation defines the anode in every kind of cell, so that part never changes. What changes is polarity: in a galvanic cell the anode is the negative terminal, but in an electrolytic cell the external supply makes it positive so that it attracts anions. Option three is the galvanic answer, offered to catch candidates who memorised "anode is negative" instead of "anode is oxidation".

  2. For Zn + Cu²⁺ → Zn²⁺ + Cu, E°_cell = 1.10 V. For 2Zn + 2Cu²⁺ → 2Zn²⁺ + 2Cu, E°_cell is:

    1. 2.20 V
    2. 1.10 V
    3. 0.55 V
    4. −1.10 V
    Show answer

    Answer: B — 1.10 V

    E° is an intensive property — energy per unit charge — so doubling the equation doubles both ΔG and n and leaves their ratio, and hence E°, unchanged at 1.10 V. Only reversing the reaction changes E°, by flipping its sign. The option 2.20 V is offered in nearly every question of this type because scaling is the instinctive move for ΔG and candidates carry the habit across.

  3. Electrolysis of a concentrated aqueous solution of NaCl gives, at the cathode:

    1. Sodium metal
    2. Hydrogen gas
    3. Chlorine gas
    4. Oxygen gas
    Show answer

    Answer: B — Hydrogen gas

    Water competes at the cathode and is reduced far more readily than Na⁺, whose E° of −2.71 V makes it the last thing to be reduced in aqueous solution, so hydrogen is evolved. Sodium metal can only be obtained from molten NaCl, where no water is present — that is the whole reason the Downs cell operates on the melt. Chlorine is the anode product, not the cathode product.

  4. One faraday of charge is passed through solutions of AgNO₃ and CuSO₄ in series. The masses deposited are:

    1. 108 g Ag and 63.5 g Cu
    2. 108 g Ag and 31.75 g Cu
    3. 54 g Ag and 31.75 g Cu
    4. Equal masses of both
    Show answer

    Answer: B — 108 g Ag and 31.75 g Cu

    One faraday deposits one gram equivalent, which is M/n. Silver is Ag⁺ with n = 1, giving 108 g; copper is Cu²⁺ with n = 2, giving 63.5/2 = 31.75 g. The first option uses the full atomic mass of copper and so ignores its charge of 2 — the commonest error in Faraday numericals, and the reason the equivalent mass rather than the molar mass must always be used.

  5. On dilution, the molar conductivity of an electrolyte:

    1. Decreases, like conductivity
    2. Increases, while conductivity decreases
    3. Remains constant
    4. Increases only for strong electrolytes
    Show answer

    Answer: B — Increases, while conductivity decreases

    Conductivity κ measures conductance per unit volume, and dilution puts fewer ions in that volume, so κ falls. Molar conductivity counts conductance per mole of electrolyte, and dilution reduces interionic attraction — and for weak electrolytes also increases dissociation — so Λ_m rises. It rises for both types, only far more steeply for weak electrolytes, which is why the last option is wrong.

  6. A galvanic cell stops working when:

    1. E° becomes zero
    2. The cell reaction reaches equilibrium, so E becomes zero
    3. All the electrolyte has evaporated
    4. The salt bridge is removed, lowering E°
    Show answer

    Answer: B — The cell reaction reaches equilibrium, so E becomes zero

    As the cell discharges, product ions build up and reactant ions deplete, so Q rises towards K and the Nernst term drags E down to zero — at which point the reaction is at equilibrium and no more work can be extracted. E° is a fixed standard-state constant and cannot become zero as the cell runs; removing the salt bridge stops the cell by breaking the circuit, but again without changing E°.

  7. Galvanising protects iron from rusting even at a scratch because zinc:

    1. Forms an impermeable physical barrier
    2. Has a more negative E° and so becomes the anode, dissolving in place of iron
    3. Is chemically inert
    4. Reacts with oxygen to give a gas that displaces air
    Show answer

    Answer: B — Has a more negative E° and so becomes the anode, dissolving in place of iron

    Zinc at −0.76 V is more easily oxidised than iron at −0.44 V, so wherever both are exposed together the zinc becomes the anode and corrodes while the iron stays cathodic and intact — cathodic protection by a sacrificial anode. A purely physical barrier, which is what paint provides, fails the moment it is scratched, and that contrast is exactly what the question is testing.

  8. The standard hydrogen electrode is assigned a potential of exactly 0.00 V because:

    1. Hydrogen genuinely has no tendency to be reduced
    2. A single electrode potential cannot be measured, so one reference must be defined
    3. Its potential happens to be zero at 298 K only
    4. Platinum is inert
    Show answer

    Answer: B — A single electrode potential cannot be measured, so one reference must be defined

    Any potential measurement needs a complete circuit and therefore two electrodes, so only differences are observable and the zero of the scale is a matter of convention, like sea level for altitude. Hydrogen does have a real tendency to be reduced — that is why metals below it in the series displace hydrogen from acids. Platinum being inert is why it is a suitable inert conductor for the electrode, not why the potential is defined as zero.