Why does the left side always show the anode?

The drawing is reorganized after reading your two half-cell choices so that oxidation (lower E° as a reduction couple) is always on the left and reduction on the right. That matches the usual textbook sketch of electron travel through the external wire from anode to cathode.

How is Q built when silver (Ag⁺/Ag, one electron) is paired with zinc (two electrons)?

The overall reaction is balanced with the least common multiple of the half-reaction electron counts. For Zn + 2Ag⁺ → Zn²⁺ + 2Ag, n = 2 and Q = [Zn²⁺]/[Ag⁺]² when solids are omitted from Q.

Why might my voltage disagree with a measured cell?

Real cells have junction potentials at the salt bridge, non-unity activity coefficients, and often side reactions or passivation. The simulator uses 1 M standard states, idealized E° values, and concentrations as activities.

What if both half-cells are the same metal?

Then E°cell = 0 and, with identical treatment of both sides, E_cell = 0—the cartoon represents no net galvanic drive.

Galvanic (Voltaic) Cell

A galvanic (voltaic) cell harnesses a spontaneous redox reaction to push electrons through an external circuit. This page shows a schematic two-beaker layout with a salt bridge, metal electrodes, and a voltmeter on the wire. You choose two metal ion / metal couples (standard reduction potentials E° vs SHE are rounded textbook values). The model automatically assigns anode (oxidation, lower E° couple) on the left and cathode (reduction, higher E° couple) on the right so the diagram always matches spontaneous electron flow left → right in the wire. Ion concentrations in each half-cell feed the overall reaction quotient Q for the balanced cell reaction (including correct stoichiometric powers when n differs between halves, e.g. Zn with Ag⁺). The Nernst equation E_cell = E°cell − (RT/nF) ln Q updates the displayed potential; temperature is adjustable. Electron dots animate along the external path when E_cell ≥ 0 and slow or reverse for non-spontaneous combinations. Junction potentials, activity coefficients, complex speciation, and concentration overpotential are omitted—concentrations stand in for activities—so numbers are for conceptual practice next to the standalone Nernst equation lab.

Who it's for: High school and introductory college chemistry alongside redox, standard potentials, and the Nernst equation; bridges formula practice to a whole-cell picture.

Key terms

Galvanic cell
Anode and cathode
Standard reduction potential
Salt bridge
Cell potential
Nernst equation
Reaction quotient
Faraday constant

How it works

A voltaic (galvanic) cell converts spontaneous redox chemistry into electrical work. This page sketches two beakers, a salt bridge, and an external wire with a voltmeter. Standard reduction potentials decide which side oxidizes; the Nernst equation updates E_cell when ion concentrations depart from 1 M.

Key equations

E°cell = E°cathode − E°anode (both as reductions vs SHE)

E_cell = E°cell − (RT / nF) ln Q

Frequently asked questions

Why does the left side always show the anode?: The drawing is reorganized after reading your two half-cell choices so that oxidation (lower E° as a reduction couple) is always on the left and reduction on the right. That matches the usual textbook sketch of electron travel through the external wire from anode to cathode.
How is Q built when silver (Ag⁺/Ag, one electron) is paired with zinc (two electrons)?: The overall reaction is balanced with the least common multiple of the half-reaction electron counts. For Zn + 2Ag⁺ → Zn²⁺ + 2Ag, n = 2 and Q = [Zn²⁺]/[Ag⁺]² when solids are omitted from Q.
Why might my voltage disagree with a measured cell?: Real cells have junction potentials at the salt bridge, non-unity activity coefficients, and often side reactions or passivation. The simulator uses 1 M standard states, idealized E° values, and concentrations as activities.
What if both half-cells are the same metal?: Then E°cell = 0 and, with identical treatment of both sides, E_cell = 0—the cartoon represents no net galvanic drive.