Electrochemical cells

An electrochemical cell turns the energy of a redox reaction into electrical energy. Two half-cells, each a metal (or platinum) dipping into a solution of ions, are joined by a wire and a salt bridge. The half-cell that releases electrons more easily becomes the negative electrode. Standard electrode potentials, E°, measured against the standard hydrogen electrode, tell you which way the electrons flow and the e.m.f. of the cell.

Key facts

  • A half-cell is written as a reduction: Zn²⁺(aq) + 2e⁻ ⇌ Zn(s). Its standard electrode potential, E°, is measured against the standard hydrogen electrode (E° = 0.00 V) at 298 K, 100 kPa and 1.00 mol dm⁻³.
  • The more negative E° is, the more easily the metal loses electrons (a stronger reducing agent). The more positive E° is, the more easily the oxidised form gains electrons (a stronger oxidising agent).
  • In a cell the half-cell with the more positive E° is the positive electrode, where reduction happens; the other is the negative electrode, where oxidation happens. Electrons flow through the wire from the negative to the positive electrode.
  • E°cell = E°(positive electrode) − E°(negative electrode). In cell notation, Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), the right-hand electrode is positive: E°cell = E°right − E°left.
  • The salt bridge (filter paper soaked in potassium nitrate solution) lets ions move between the half-cells to balance the charge, completing the circuit.
  • A reaction is feasible if E°cell is positive: an oxidising agent on the left of one half-equation reacts with a reducing agent on the right of a half-equation with a less positive E°.
  • Non-rechargeable cells stop when a reactant is used up; rechargeable cells such as lithium-ion cells are reversed by charging; fuel cells are fed with fuel, and a hydrogen–oxygen fuel cell makes only water.

Worked examples

Example 1

A cell is made from a zinc half-cell (E° = −0.76 V) and a copper half-cell (E° = +0.34 V). Find E°cell, say which electrode is positive and write the overall equation.

  1. Copper has the more positive E°, so Cu²⁺ ions gain electrons there: copper is the positive electrode.
  2. Zinc has the more negative E°, so zinc atoms lose electrons: zinc is the negative electrode.
  3. Find the e.m.f. E°cell = +0.34 − (−0.76) = 1.10 V
  4. Reverse the zinc half-equation and add it to the copper one; the 2 electrons cancel. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Answer: E°cell = 1.10 V; copper is the positive electrode; Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Example 2

Cl₂ + 2e⁻ ⇌ 2Cl⁻ E° = +1.36 V; Br₂ + 2e⁻ ⇌ 2Br⁻ E° = +1.09 V; I₂ + 2e⁻ ⇌ 2I⁻ E° = +0.54 V. Will chlorine oxidise bromide ions? Will iodine oxidise bromide ions?

  1. Chlorine (left of its half-equation) can oxidise anything on the right of a half-equation with a less positive E°. +1.36 > +1.09
  2. So Cl₂ oxidises Br⁻ to Br₂. E°cell = 1.36 − 1.09 = +0.27 V
  3. For iodine the E° of I₂/I⁻ is less positive than that of Br₂/Br⁻. E°cell = 0.54 − 1.09 = −0.55 V
  4. E°cell is negative, so iodine does not oxidise bromide ions. This matches the order of reactivity of the halogens.

Answer: Chlorine will (E°cell = +0.27 V); iodine will not (E°cell = −0.55 V).

Example 3

Write the overall equation and find E°cell for a cell made from Al³⁺/Al (E° = −1.66 V) and Ag⁺/Ag (E° = +0.80 V).

  1. Aluminium has the more negative E°, so it is oxidised: Al(s) → Al³⁺(aq) + 3e⁻.
  2. Silver ions are reduced. Each takes one electron, so three are needed. 3Ag⁺(aq) + 3e⁻ → 3Ag(s)
  3. Add and cancel the electrons. Al(s) + 3Ag⁺(aq) → Al³⁺(aq) + 3Ag(s)
  4. E° values are never multiplied by the numbers in the equation. E°cell = 0.80 − (−1.66) = 2.46 V

Answer: Al(s) + 3Ag⁺(aq) → Al³⁺(aq) + 3Ag(s); E°cell = 2.46 V