Galvanic cells

A galvanic cell does not create a new kind of chemistry. It takes a spontaneous redox reaction and separates oxidation from reduction so that electrons are forced through an external circuit, where their flow can do electrical work.

The redox reaction is spatially separated. Electrons travel through the external circuit while ions move through the electrolyte connection so charge does not accumulate in either half-cell.

Build the cell from two half-reactions

The Daniell cell provides a clean example:

Zn(s) → Zn²⁺(aq) + 2e⁻ oxidation

Cu²⁺(aq) + 2e⁻ → Cu(s) reduction

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) overall

Oxidation occurs at the anode; reduction occurs at the cathode. In a galvanic cell, the spontaneous reaction makes the anode negative relative to the cathode, so electrons move through the wire from Zn toward Cu.

The salt bridge completes the ionic circuit

At the zinc electrode, Zn²⁺ enters solution. At the copper electrode, Cu²⁺ leaves solution as Cu(s). If nothing compensated for these changes, charge separation would rapidly oppose further electron transfer.

The salt bridge or porous separator lets ions migrate so that each half-cell remains approximately electroneutral. In the simple Daniell picture, anions move toward the anode compartment and cations toward the cathode compartment.

The bridge does not carry electrons between the electrodes. Electrons use the external metallic circuit; ions carry current through the electrolyte.

Cell notation

A compact representation is:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

A single vertical line marks a phase boundary. The double separator represents the liquid junction or salt-bridge connection. Reading the notation from left to right should reproduce the physical sequence of phases across the cell.

Cell potential and spontaneity

Standard electrode potentials are tabulated as reduction potentials. For a galvanic cell:

E°_cell = E°_cathode − E°_anode

Using E°(Cu²⁺/Cu) ≈ +0.34 V and E°(Zn²⁺/Zn) ≈ −0.76 V gives:

E°_cell = 0.34 − (−0.76) = +1.10 V

A positive E_cell corresponds to a spontaneous cell reaction under the stated conditions. The thermodynamic relation is ΔG = −nFE_cell.

Do not multiply an electrode potential by a stoichiometric coefficient. Potential is intensive: balancing electrons changes the amount of reaction, not the voltage assigned to a half-cell.

Positive cell potential and negative Gibbs free-energy change describe the same spontaneous direction. For the cell reaction, ΔG = −nFE.

Composition changes the voltage

Standard potential refers to standard-state conditions. Away from them, composition changes the driving force through the Nernst equation:

E = E° − (RT/nF) ln Q

For the Daniell reaction, increasing Zn²⁺ or decreasing Cu²⁺ increases Q and lowers E. As equilibrium is approached, the chemical driving force falls and E tends toward zero.

This is the electrical expression of the same thermodynamics that governs reaction free energy.

Equilibrium potential and operating voltage

Thermodynamic potential

Sets the reversible driving voltage for the chemical state.

Terminal voltage under load

Also reflects internal resistance, charge-transfer kinetics and mass transport.

A cell can have a favourable positive equilibrium potential yet deliver poor power if reactions are slow or internal transport is resistive. Voltage and power are therefore related but not interchangeable measures of cell performance.

Galvanic and electrolytic cells

Anode

Oxidation occurs here.

Galvanic cell: normally negative.

Cathode

Reduction occurs here.

Galvanic cell: normally positive.

In an electrolytic cell, an external power supply drives a nonspontaneous reaction and the electrode signs reverse. The chemical definitions do not: anode still means oxidation and cathode still means reduction.

Exercises

Identify the electrodes

Mg(s) becomes Mg²⁺ while Ag⁺ becomes Ag(s). Which electrode is the anode, and which way do electrons move?

Solution

Mg is oxidized, so Mg is the anode. Electrons move through the external circuit from Mg toward the Ag cathode.

Standard cell potential

Use E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁺/Cu) = +0.34 V. Cu is oxidized and Ag⁺ reduced. Find E°_cell.

Solution

E°_cell = 0.80 − 0.34 = +0.46 V.

Salt bridge

Why does current collapse if the salt bridge is removed even though the metal wire still connects the electrodes?

Solution

Charge separation builds in the half-cells. Without ionic migration to maintain approximate electroneutrality, that electrical imbalance opposes further electron transfer.