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CHEMSIM v1.0
⚡ Galvanic & Electrolytic Cells
#21

§1 Interactive Simulation

Galvanic Cell
Electrolytic Cell
3D Daniell Cell
Half-Cell Diagram
Salt Bridge Detail
⚡ THREE.JS 3D CELL
E vs Time
E° Reduction Series
ΔG vs E
Q vs E (Nernst preview)
Cell Discharge Curve
E°ₒₑₗₗ
—
Eₒₑₗₗ
—
ΔG
—
log K
—
Q
—
Spont?
—
Electron flow
Ion migration
Half-rxn labels
Voltmeter

§2 The Idea, Step by Step

🔋 From a Tug-of-War over Electrons to Cell Voltage

A battery is really just two different metals having a tug-of-war over electrons. Some metals cling to their electrons; others let go easily. Stand two of them in salty water, connect them with a wire, and the metal that lets go pushes electrons through the wire to the metal that grabs them. That moving stream of electrons is electric current — the same reason a lemon with a copper coin and a zinc nail can faintly light an LED, and the same reason you feel a tingle if you bite aluminium foil with a metal filling in your tooth.

To put numbers on it, every metal is given a score for how badly it wants electrons, called its standard reduction potential $E^\circ$, measured in volts. The metal with the higher (more positive) score wins the pull and becomes the cathode (electrons flow in, reduction happens). The loser gives up electrons and becomes the anode (oxidation). The voltage the cell produces is simply the gap between the two scores:

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ For the classic zinc–copper "Daniell" cell, zinc scores $-0.76$ V and copper scores $+0.34$ V, so $E^\circ_{\text{cell}} = 0.34 - (-0.76) = 1.10$ V — about the push of one rechargeable AA cell.

That voltage is a measure of usable energy. Moving $n$ moles of electrons across the voltage gap releases an amount of free energy $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, where $F = 96485$ C/mol is Faraday's constant. A positive $E^\circ$ gives a negative $\Delta G^\circ$, which is exactly what "the reaction runs by itself" means. Voltage isn't fixed forever, though: as the reaction uses up reactants and piles up products, the Nernst correction $E = E^\circ - \tfrac{RT}{nF}\ln Q$ pulls the working voltage down. When the products build up enough that $Q$ reaches the equilibrium constant $K$, the term cancels $E^\circ$ exactly, $E = 0$, and the battery is "dead." In the sim, the preset picks the two metals (and therefore $E^\circ$), the two concentration sliders change $Q$, the temperature slider scales the $RT/nF$ term, and the external-resistance slider sets how fast current flows.

Try this in the sim above: (1) Start on the Daniell preset and read $1.10$ V, then switch to the Ag–Zn preset and watch $E^\circ_{\text{cell}}$ jump to about $1.56$ V — silver pulls electrons harder than copper. (2) Drag [Cathode ion] down toward $0.001$ M and watch $E_{\text{cell}}$ fall below $E^\circ_{\text{cell}}$ — that is the Nernst $\ln Q$ term biting. (3) Click the Electrolytic Cell tab and see the electrons reverse direction: now an outside battery is forcing the reaction uphill ($E < 0$, $\Delta G > 0$).

§3 Equation Derivation

⚡ Standard Cell Potential & Gibbs Free Energy Relation

$$\boxed{\;E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\;}\qquad\boxed{\;\Delta G^\circ = -nFE^\circ_{\text{cell}}\;}$$

Symbol Definitions

SymbolMeaningSI Unit
$E^\circ_{\text{cell}}$Standard cell potentialV (volt)
$E^\circ_{\text{cathode/anode}}$Standard reduction potential at electrodeV
$\Delta G^\circ$Standard Gibbs free energy changeJ/mol
$n$Number of moles of electrons transferredmol
$F$Faraday's constant96 485 C/mol
$K$Equilibrium constantdimensionless
$Q$Reaction quotientdimensionless
$R$Universal gas constant8.314 J·K⁻¹·mol⁻¹

Step-by-Step Derivation

1Half-cell convention. A standard hydrogen electrode (SHE: 2H⁺ + 2e⁻ → H₂) is assigned $E^\circ = 0.000$ V at 25 °C, 1 atm, [H⁺] = 1 M. All other reduction potentials are measured relative to SHE.
2Cell potential from two half-reactions. Both standard potentials are written as reductions (gain of e⁻). The species more easily reduced becomes the cathode; the other is reversed (oxidation) to form the anode: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ Subtracting (not adding) is critical because both half-reactions are tabulated as reductions.
3Electrical work & Gibbs energy. The maximum non-PV work obtainable from a reversible cell is the electrical work moving $n$ moles of electrons across potential $E$: $$w_{\text{elec,max}} = -nFE_{\text{cell}}$$ By the second law, $\Delta G = w_{\text{non-PV,max}}$, hence: $$\boxed{\Delta G^\circ = -nFE^\circ_{\text{cell}}}$$ Sign: positive $E^\circ$ ⇒ negative $\Delta G^\circ$ ⇒ spontaneous as written.
4Link to equilibrium. At equilibrium $\Delta G^\circ = -RT\ln K$, so combining: $$-nFE^\circ_{\text{cell}} = -RT\ln K\;\Longrightarrow\;\boxed{\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}}$$ At 298 K: $\log K = \dfrac{nE^\circ_{\text{cell}}}{0.0592}$.
5Non-standard conditions (Nernst preview). When $Q \neq 1$, the cell potential is corrected: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF}\ln Q$$ This is the Nernst equation, derived in detail in topic #22.
6Power & current. For a cell connected through external resistance $R_{\text{ext}}$ with internal resistance $r$: $$I = \frac{E_{\text{cell}}}{R_{\text{ext}} + r},\qquad P_{\text{ext}} = I^2 R_{\text{ext}}$$ Maximum power transfer when $R_{\text{ext}} = r$.

Simulation Variable Mapping

UI ControlSymbolEffect
Preset selector$E^\circ_{\text{cathode}}, E^\circ_{\text{anode}}$Sets standard potentials of half-cells
[Anode ion]$Q$ denominatorHigher [anode ion] → smaller Q → higher E
[Cathode ion]$Q$ numeratorHigher [cathode ion] → larger Q → lower E
Temperature$T$ in Nernst termModifies (RT/nF) coefficient
External R$R_{\text{ext}}$Sets current flow rate, controls discharge speed

Worked Example — Daniell Cell

Cell: Zn(s)|Zn²⁺(1 M)||Cu²⁺(1 M)|Cu(s)
Half-reactions (as reductions):
Cathode: Cu²⁺ + 2e⁻ → Cu, $E^\circ = +0.34$ V
Anode (reversed): Zn → Zn²⁺ + 2e⁻, $E^\circ_{\text{red}} = -0.76$ V
$$E^\circ_{\text{cell}} = 0.34 - (-0.76) = +1.10\text{ V}$$ $$\Delta G^\circ = -nFE^\circ = -(2)(96485)(1.10) = -212.3\text{ kJ/mol}$$ $$\log K = \frac{(2)(1.10)}{0.0592} = 37.16\;\Longrightarrow\;K \approx 1.4\times 10^{37}$$ A huge K — Zn essentially fully displaces Cu²⁺, as observed.

📚 Atkins & de Paula — Physical Chemistry, 11th Ed., §6E: "Equilibrium electrochemistry" | Skoog, West, Holler & Crouch — Fundamentals of Analytical Chemistry, 9th Ed., Ch. 18: "Introduction to Electrochemistry"

§4 Frequently Asked Questions

📚 LibreTexts Chemistry — "Galvanic Cells" (chem.libretexts.org) | Khan Academy — Electrochemistry | MIT OCW 5.111

§5 Common Misconceptions

📚 Garnett & Treagust — J. Chem. Educ. 69, 121 (1992) "Conceptual difficulties in electrochemistry" | Sanger & Greenbowe — J. Chem. Educ. 74, 819 (1997) "Common student misconceptions in electrochemistry" | Taber — Chemical Misconceptions (RSC, 2002)