Why does the left side always show the anode?

The drawing is reorganized after reading your two half-cell choices so that oxidation is always on the left and reduction on the right. For dissimilar metals that means the lower-E° couple; for a concentration cell it means the more dilute ion side. 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?

That is a concentration cell: E°cell = 0, but E_cell = −(RT/nF) ln([Mⁿ⁺]_anode/[Mⁿ⁺]_cathode). The dilute side is the anode (metal dissolves); the concentrated side is the cathode (metal plates out). Equal concentrations give E_cell = 0.

Galvanic (Voltaic) Cell

Two half-cells, salt bridge, voltmeter; E°cell and Nernst E from ion concentrations (pedagogical E°).

Galvanic cell

Pick two metal couples (standard reduction potentials). The simulator assigns anode (oxidation) and cathode (reduction). Same metal + different [ion] is a concentration cell. Concentrations enter Q; E_cell follows the Nernst equation for the balanced overall reaction.

The [ion] sliders follow half-cell A and B from the menus — not necessarily the left beaker vs right. Q maps each concentration to the correct anode/cathode chemistry. Left beaker is always the anode.

Half-cell A

Half-cell B

[ion] in A (mol/L)1 M

[ion] in B (mol/L)1 M

T (K)298.15

Cell reaction

Zn + Cu²⁺ → Zn²⁺ + Cu

Q = [Zn²⁺] / [Cu²⁺]

Presets

Shortcuts

R — reset to Zn | Cu, 1 M, 298 K

Measured values

AnodeZn

CathodeCu

E°cell1.100 V

E_cell1.1000 V

Q1.000e+0

n (e⁻ in overall)2

About this model

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) on the left and cathode (reduction) on the right so the diagram matches spontaneous electron flow left → right in the wire. For dissimilar metals the anode is the lower-E° couple; for the same metal with different ion concentrations the dilute side is the anode (concentration cell). Ion concentrations feed the overall reaction quotient Q (including correct stoichiometric powers when n differs, e.g. Zn with Ag⁺). The panel shows the balanced cell reaction and symbolic Q. The Nernst equation E_cell = E°cell − (RT/nF) ln Q updates the potential; temperature is adjustable. Electron dots animate on the external wire when |E_cell| is appreciable; the salt bridge shows schematic cation → cathode and anion → anode arrows. 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
Concentration cell
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. Same metal with different concentrations is a concentration cell.

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 is always on the left and reduction on the right. For dissimilar metals that means the lower-E° couple; for a concentration cell it means the more dilute ion side. 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?: That is a concentration cell: E°cell = 0, but E_cell = −(RT/nF) ln([Mⁿ⁺]_anode/[Mⁿ⁺]_cathode). The dilute side is the anode (metal dissolves); the concentrated side is the cathode (metal plates out). Equal concentrations give E_cell = 0.

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