Galvanic cells, electrode potentials and batteries

Chemistry · Electrochemistry. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A galvanic (voltaic) cell turns the energy of a spontaneous redox reaction into electricity. Oxidation and reduction happen in separate half-cells, so the electrons must travel through the outer wire. A salt bridge joins the two solutions.

Each half-cell has a standard electrode potential, E°, measured against the standard hydrogen electrode. The list of E° values in order is the electrochemical series. A metal with a more negative E° loses electrons more easily.

The half-cell with the lower (more negative) reduction potential becomes the anode. The cell emf is the difference between the two potentials. A positive emf means the reaction is feasible.

Rules to remember

Galvanic cell: anode is negative, oxidation occurs; cathode is positive, reduction occurs. Electrons flow from anode to cathode in the wire.
Daniell cell: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s); zinc is the anode, copper the cathode, emf 1.10 V.
Salt bridge: completes the circuit and keeps both solutions electrically neutral by letting ions move.
Standard hydrogen electrode: platinised platinum, H₂ at 1 atm, 1 mol dm⁻³ H⁺, 25 °C; its potential is taken as 0.00 V.
E°_cell = E°(cathode) − E°(anode), using standard reduction potentials for both.
A metal displaces from solution the ions of any metal below it in the series (one with a less negative E°). Metals above hydrogen displace H₂ from acids.
Lead accumulator (rechargeable, 2 V per cell): Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O on discharge. Alkaline and silver oxide cells use a zinc anode with KOH; a hydrogen-oxygen fuel cell gives only water.

Common mistake

The signs of the electrodes are opposite in the two kinds of cell. In a galvanic cell the anode is negative; in an electrolytic cell the anode is positive. In both, oxidation happens at the anode.

Worked example

E°(Fe²⁺/Fe) = −0.44 V and E°(Cu²⁺/Cu) = +0.34 V. Which metal is the anode in a cell made from these half-cells, and what is the emf?

Iron has the more negative reduction potential, so iron is oxidised: iron is the anode.
E°_cell = E°(cathode) − E°(anode) = (+0.34) − (−0.44)
E°_cell = 0.34 + 0.44 = 0.78 V
The emf is positive, so Fe + Cu²⁺ → Fe²⁺ + Cu is feasible.

Answer: Iron is the anode (negative electrode) and the emf is 0.78 V.

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