A Daniell cell, discharged through a load

Zn | ZnSO₄ | salt bridge | CuSO₄ | Cu, with every ion simulated, the bridge's included: the voltage across the load, the electrode levels that set it, the salt leaking out of the bridge, and where the voltage goes. Choose Daniell's own porous pot instead of the bridge, a lighter or heavier load, and how fast to play it. (More about the model, and what to look for, below.)

About this simulation

The cell is built in a tube: the two solutions are each 4 cm long and stirred, the salt bridge 2 cm of unstirred 3 M KCl, and everything per unit area of the tube's bore (a 1D model). A real cell's beakers are far bigger than its bridge's bore, so their solutions change more slowly than here. Every ion is present everywhere, if only as a 1 µM trace, so each can cross every junction; Cu²⁺ in the zinc's solution starts negligible and grows only as copper seeps across. Each end of the bridge starts as a thin zone where the two solutions are pre-mixed, as at a frit (a perfectly sharp start is harder for the solver, and no real junction is sharp).

The first diagram is the electrons' view, in volts. Each metal's Fermi level is the voltage of its electrons. Each couple (Zn²⁺/Zn, Cu²⁺/Cu) has a redox level: the electron voltage at which the local solution and metal would be in equilibrium, so a metal sitting at its couple's level exchanges no net current. At open circuit the two Fermi levels sit at their couples' levels, 1.10 V apart. Under load each metal departs from its couple's level by an overpotential, the levels tilt across the solutions (ohmic drop, and composition changes), and what's left reaches the load. The dashed thin lines are the couples' standard levels, which ride on the SHE's standard level (φ itself, with table data): zinc's 0.76 V below it, copper's 0.34 V above.

The ion diagram uses the species voltage V_i = μ̄_i/(z_iF): an ion's electrochemical potential expressed in volts per unit charge, so it can be drawn on the same axis as the electrons' Fermi level. Each ion moves down its own μ̄_i, so its current flows down slopes in its own V_i (cations move downhill, anions uphill), and an ion in equilibrium with a metal's electrons sits level with them (Zn²⁺ with zinc's electrons, Cu²⁺ with copper's). The thin line under each, the standard level V°_i, is where V_i would be at the reference concentration; the gap between them is the concentration term. The ions' levels sit volts apart, so each is drawn with a display offset (⌇) to share one plot.

The bridge is where a textbook says "assume the liquid junction potentials are negligible". Here they're simulated: at open circuit, φ steps by +0.97 mV at the zinc side's junction and −0.97 mV at the copper side's (Henderson's formula gives +0.99 and −1.00 mV for these solutions), nearly cancelling, because K⁺ and Cl⁻ move at nearly the same speed. Meanwhile every ion's own voltage changes by tens of mV across each junction, and that drives the leak: KCl spreads out into the solutions, and Zn²⁺, Cu²⁺ and SO₄²⁻ into the bridge. Over hours the leaked KCl makes the solutions more conductive, so under a heavy load the voltage rises at first. Once the KCl has spread everywhere, what's left of the junction is Zn²⁺ and Cu²⁺ diffusing past each other in a KCl background: with nearly equal mobilities and that supporting electrolyte, its potential stays tiny. (The liquid-junction demo takes junctions on their own: Henderson's formula against Planck's and a junction growing from contact, and what a charged frit does.)

Daniell's own separator was an unglazed earthenware pot: a porous wall that passes every ion, only slowly (here 5 mm of pores, where ions diffuse ten times slower than in free solution). There's no KCl to leak, and the junction between the two sulfate solutions is tiny. Either way, though, copper sulfate seeps across, and Cu²⁺ that reaches the zinc plates straight onto it, a short circuit inside the cell that wastes its charge. The status line counts it (as the copper that's gone without passing through the load), and the slower the discharge, the bigger its share: at the lightest load here, over ten days, the bridged cell loses about a fifth of its charge this way and the potted one about a twentieth.