§ C-8 Daniell–Jacobi Galvanic Cell & Electrode Potentials (Grade 9–11 · Chemistry)
Anodic oxidation Zn⁰ → Zn²⁺ + 2e⁻, cathodic reduction Cu²⁺ + 2e⁻ → Cu⁰ & salt bridge. A spatially separated redox reaction converts the chemical Gibbs free energy change ΔG directly into an ordered flow of electrons, electric current.
Theory Summary & Scientific Mechanism
The Daniell galvanic cell converts the chemical free energy of the spontaneous redox reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) directly into electrical work by spatially separating the half-reactions. At the anode (negative terminal), zinc metal undergoes oxidation and dissolves: Zn⁰ - 2e⁻ → Zn²⁺ (E^° = -0.76 V). Electrons travel through the external circuit to the cathode (positive terminal), where copper ions are reduced and plate out: Cu²⁺ + 2e⁻ → Cu⁰ (E^° = +0.34 V).
The cell electromotive force (EMF) at arbitrary ion concentrations is governed by the Nernst equation: E_cell = E^°_cell - (RT)/(zF)ln ([Zn²⁺])/([Cu²⁺]), where E^°_cell = E^°_cathode - E^°ₐnode = +0.34 - (-0.76) = 1.10 V and z = 2. The maximum electrical work equals the decrease in Gibbs free energy: Wₑlec,max = -Δ G = z F E_cell.
The salt bridge (a U-tube filled with KCl or KNO₃ gel) is essential to close the internal circuit: without it, excess Zn²⁺ charge in the anode half-cell and uncompensated SO₄²⁻ anions in the cathode half-cell would build an opposing liquid-junction potential that halts electron flow within microseconds.
Everyday Analogy
The Daniell cell is the direct ancestor of every smartphone and EV battery! More reactive zinc Zn eagerly gives up electrons to Cu²⁺ ions, and because the two half-reactions are separated into two beakers, those electrons must travel through the external wire to do electrical work.
Interactive Question & Derivation
In which direction do electrons flow through the external wire of a Daniell galvanic cell, and what function does the KCl salt bridge perform?
- A: Electrons flow from the more active zinc anode (E^° = -0.76 V) to the copper cathode (E^° = +0.34 V), while the salt bridge maintains electrical neutrality by migrating K⁺ and Cl⁻ ions
- B: Free electrons swim through the aqueous salt bridge from copper to zinc
- C: The zinc electrode acts as the cathode and gains mass during discharge
Study Hint: Compare standard potentials: E^°(Zn²⁺/Zn) = -0.76 V < E^°(Cu²⁺/Cu) = +0.34 V. Which metal oxidizes as the anode, which way do electrons flow, and what is E^°_cell = E^°_cathode - E^°ₐnode?
Proof & Explanation: Zinc has a more negative standard reduction potential (E^° = -0.76 V) than copper (E^° = +0.34 V). Oxidation occurs at the zinc anode (Zn⁰ → Zn²⁺ + 2e⁻) and reduction at the copper cathode (Cu²⁺ + 2e⁻ → Cu⁰). By the Nernst equation, higher [Cu²⁺] and lower [Zn²⁺] increase cell EMF E.
Key Equations & 3D Simulation Parameters
E = 1.100 - (8.314 · T)/(2 · 96485)ln ([Zn²⁺])/([Cu²⁺]), Δ G = -z F E
- Catholyte concentration [Cu²⁺] (CuSO₄): 0.05 … 2 mol/L (default: 1 mol/L)
- Anolyte concentration [Zn²⁺] (ZnSO₄): 0.05 … 2 mol/L (default: 1 mol/L)
- Cell temperature T: 273 … 353 K (default: 298 K)
- External load conductance (discharge rate): 0.4 … 2 × (default: 1 ×)