CHEMISTRY INTERMEDIATE

Electrochemistry

Electrons moving through circuits and through solutions — how spontaneous redox reactions power batteries, how electrolysis uses electricity to drive chemical change, and how the Nernst equation quantifies it all.

01 · Foundations

Redox reactions and oxidation states

Electrochemistry is built on redox reactions — processes in which electrons are transferred between species. Oxidation is the loss of electrons; reduction is the gain. The two always occur together: the species that loses electrons (the reducing agent) causes another species to gain them (the oxidising agent).

OXIDATION

Loss of electrons. Oxidation state increases. The reducing agent is oxidised.

Zn(s) → Zn²⁺(aq) + 2e⁻

REDUCTION

Gain of electrons. Oxidation state decreases. The oxidising agent is reduced.

Cu²⁺(aq) + 2e⁻ → Cu(s)

Oxidation states are bookkeeping numbers that track electron ownership in a compound. Rules: free elements are 0; ions are equal to their charge; oxygen is usually −2; hydrogen is usually +1. For species in solution, balance atoms, then balance charge with electrons to write a half-equation.

For practice balancing redox equations, see the Balancing equations guide.

02 · Galvanic cells

Turning spontaneous reactions into electricity

A galvanic cell (voltaic cell) harnesses a spontaneous redox reaction to produce an electric current. The classic Daniell cell separates the Zn/Zn²⁺ and Cu/Cu²⁺ half-cells:

ANODE (−)

Zn → Zn²⁺ + 2e⁻

Oxidation occurs here. Zinc electrode dissolves.

salt bridge

⇌ ions ⇌

CATHODE (+)

Cu²⁺ + 2e⁻ → Cu

Reduction occurs here. Copper deposits.

The cell notation for the Daniell cell is:

Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)

The cell potential under standard conditions is:

E°cell = E°cathode − E°anode

E°cell = +0.34 − (−0.76) = +1.10 V

A positive E°cell indicates that the reaction is spontaneous under standard conditions (ΔG° = −nFE°cell).

Standard electrode potential table

Half-reaction (reduction) E° / V
F₂(g) + 2e⁻ → 2F⁻(aq) +2.87
MnO₄⁻(aq) + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O +1.51
Cl₂(g) + 2e⁻ → 2Cl⁻(aq) +1.36
O₂(g) + 4H⁺ + 4e⁻ → 2H₂O(l) +1.23
Fe³⁺(aq) + e⁻ → Fe²⁺(aq) +0.77
Cu²⁺(aq) + 2e⁻ → Cu(s) +0.34
2H⁺(aq) + 2e⁻ → H₂(g) 0.00
Fe²⁺(aq) + 2e⁻ → Fe(s) −0.44
Zn²⁺(aq) + 2e⁻ → Zn(s) −0.76
Li⁺(aq) + e⁻ → Li(s) −3.04

Species at the top of the table are strong oxidising agents (high tendency to be reduced). Species at the bottom are strong reducing agents (high tendency to be oxidised). A spontaneous galvanic cell always has the stronger oxidising agent at the cathode.

03 · Non-standard conditions

The Nernst equation

Standard potentials apply only at 1 mol/L, 298 K, and 100 kPa. At other concentrations or temperatures, the cell potential changes. The Nernst equation makes this quantitative:

E = E° − (RT / nF) ln Q

E Cell potential at non-standard conditions (V)
E° Standard cell potential (V)
R Gas constant = 8.314 J mol⁻¹ K⁻¹
T Temperature (K)
n Moles of electrons transferred in the balanced equation
F Faraday constant = 96 485 C mol⁻¹
Q Reaction quotient (products/reactants, each raised to stoichiometric power)

At 298 K, the Nernst equation simplifies to:

E = E° − (0.0592 / n) log Q

When the cell reaches equilibrium, E = 0 and Q = K (the equilibrium constant). Rearranging gives the important relationship:

ln K = nFE° / RT

This connects electrochemistry to thermodynamics and equilibrium — topics explored further in the Reaction rates guide and the Acids and bases guide (Ka and Kb are equilibrium constants of the same form).

04 · Electrolytic cells

Electrolysis and Faraday's laws

An electrolytic cell uses an external power supply to drive a non-spontaneous redox reaction. The positive terminal of the supply connects to the anode; the negative terminal to the cathode — the opposite of a galvanic cell's spontaneous convention for sign.

Key industrial applications:

  • Aluminium smelting — electrolysis of molten Al₂O₃ (Hall–Héroult process). Al³⁺ is reduced at the cathode; O²⁻ is oxidised at the carbon anode.
  • Chlorine production — electrolysis of brine (NaCl(aq)) produces Cl₂ at the anode, H₂ at the cathode, and NaOH in solution (chloralkali process).
  • Electroplating — the object to be plated is the cathode; the plating metal is the anode. Metal ions in solution deposit onto the object.
  • Electrolytic refining of copper — impure copper dissolves at the anode; pure copper deposits at the cathode. Precious metal impurities fall as anode slime.

Faraday's law calculation

m = (M × I × t) / (n × F)

mMass deposited or dissolved (g)
MMolar mass of the substance (g/mol) — see Molar mass
ICurrent (amperes)
tTime (seconds)
nElectrons transferred per ion (from the half-equation)
FFaraday constant = 96 485 C mol⁻¹

Worked example: A current of 2.0 A passes through a copper sulfate solution for 30 minutes. How much copper deposits?

Cu²⁺ + 2e⁻ → Cu (so n = 2)

m = (63.5 × 2.0 × 1800) / (2 × 96 485)

m = 228 600 / 192 970 ≈ 1.18 g

Electrochemistry connects directly to nuclear fusion — the fuel-cycle analysis requires understanding of how tritium (the fusion fuel) is produced by neutron bombardment, an analogue of electrolytic reduction. And in biology, the electron transport chain in mitochondria is essentially a biological galvanic cell: electrons flow from NADH (reducing agent) to oxygen (oxidising agent) through a series of redox-active proteins. See Cellular respiration for the biochemical version.

Electrochemistry — common questions

Oxidation is the loss of electrons; reduction is the gain of electrons. The two always occur together — one species is oxidised while another is reduced. A related concept is oxidation state, which tracks the apparent charge on an atom in a compound. See also Balancing equations for redox balancing.

Master the calculations

The Molar mass calculator lets you work out M for any formula — the first number you need in every Faraday's law problem.

Molar mass calculator →