Two electrodes, one spontaneous reaction, a wire between them
A galvanic (voltaic) cell converts a spontaneous redox reaction directly into electrical current by physically separating the two half-reactions. At the anode, a metal loses electrons and dissolves into solution as ions — oxidation. Those electrons travel through an external wire (doing useful work along the way) to the cathode, where a different ion in solution gains them and plates out as metal — reduction. A salt bridge completes the circuit internally, letting ions migrate to keep both half-cells electrically neutral without letting the two solutions mix directly and react uselessly on contact.
The Daniell cell: the textbook example
John Daniell's 1836 design pairs a zinc electrode in zinc sulfate with a copper electrode in copper sulfate. Zinc is more easily oxidised than copper, so the reaction runs spontaneously in one direction only:
anode (oxidation): Zn(s) -> Zn2+(aq) + 2e- cathode (reduction): Cu2+(aq) + 2e- -> Cu(s) overall: Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s) E-cell(standard) = +1.10 V
Standard electrode potentials
Every half-reaction has a standard reduction potential, E°, measured against a defined reference (the standard hydrogen electrode, fixed at 0 V by convention) under standard conditions of 1 M concentration, 1 atm, 25°C. A cell's standard voltage is simply the cathode's E° minus the anode's E°: E°cell = E°cathode − E°anode. Whichever half-reaction has the higher (more positive) reduction potential wins and runs as written — as reduction, at the cathode — while the other is forced to run backward, as oxidation, at the anode. This single comparison is why a table of standard electrode potentials lets you predict, without running the experiment, which of any two metals will end up dissolving and which will end up plating.
The Nernst equation: when concentrations aren't standard
Real cells are rarely at exactly 1 M concentration, and cell voltage depends on concentration because dissolving more product ions makes the reaction less favourable (Le Chatelier), while starting with more reactant ions makes it more favourable. The Nernst equation corrects the standard potential for the actual concentrations in play:
E = E_standard - (R*T)/(n*F) * ln(Q) R = gas constant, T = temperature (K), n = electrons transferred, F = Faraday constant Q = reaction quotient = [products] / [reactants] (at that instant, not at equilibrium) at 25 C this simplifies to: E = E_standard - (0.0592/n) * log10(Q)
That 0.0592/n V factor is small on paper, but it explains real, measurable behaviour: as a Daniell cell discharges, Zn²⁺ builds up and Cu²⁺ is depleted, Q rises, and the cell voltage drifts measurably below its standard 1.10 V long before either electrode is anywhere near fully consumed — exactly what you would notice as a battery's voltage sagging under prolonged use.
Why the salt bridge is not optional
Without a path for ions to migrate between the two half-cells, oxidation at the anode would leave that solution with a runaway positive charge (excess Zn²⁺) while reduction at the cathode would leave its solution increasingly negative (depleted Cu²⁺ relative to its sulfate), and both effects would very quickly stop the reaction cold — an unbalanced charge opposes further electron flow far more strongly than the tiny Nernst correction does. The salt bridge, typically an inert electrolyte like KCl in a gel, lets K⁺ drift toward the cathode compartment and Cl⁻ toward the anode compartment, neutralising the buildup on both sides without letting Cu²⁺ and Zn metal ever touch and react directly (which would just waste the electrons as heat instead of routing them through the external circuit).
Voltage, free energy and how much work you actually get
Cell potential connects directly to thermodynamics through ΔG = −nFE: a positive cell voltage corresponds to a negative Gibbs free energy change, confirming the reaction is spontaneous, and the magnitude tells you the maximum electrical work extractable per mole of reaction — n is the moles of electrons transferred and F is Faraday's constant, the charge on one mole of electrons (≈96,485 C/mol). This is the thermodynamic bridge between chemistry and electrical engineering: it's the same equation, differently dressed, whether you're describing a chemistry-class Daniell cell or the lithium-ion cell in a phone battery.
Frequently asked questions
What is the actual difference between the anode and the cathode?
The anode is where oxidation happens — a species loses electrons — and the cathode is where reduction happens — a species gains them. In a galvanic cell the anode is where the metal dissolves and electrons enter the external circuit; that assignment flips in an electrolytic cell, where an external power source drives the reaction backward.
Why does a battery's voltage sag before it's fully discharged?
As the reaction proceeds, reactant concentrations fall and product concentrations rise, which the Nernst equation translates directly into a lower cell voltage even though the standard potential hasn't changed. The chemistry hasn't run out — the concentrations have simply drifted away from the standard-state conditions the nominal voltage assumes.
What would happen if you removed the salt bridge from a galvanic cell?
Charge would build up almost immediately in both half-cells — excess positive ions at the anode, a relative deficit at the cathode — and that unbalanced charge would very quickly oppose and stop further electron flow, since nothing is available to neutralise it internally. The circuit effectively dies within moments, well before the reactants are used up.
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