Lesson 18: Galvanic and Electrolytic CellsAP

An electrochemical cell converts between chemical and electrical energy. A galvanic (voltaic) cell harvests a spontaneous redox reaction as electricity: the two half-reactions are separated into half-cells, electrons are forced to travel through an external wire from the species being oxidised to the species being reduced, and a salt bridge (or porous barrier) lets ions migrate to keep each half-cell electrically neutral. The electrode where oxidation occurs is the anode — in a galvanic cell it is the negative terminal, because electrons pile up there as they are produced. Reduction happens at the cathode, the positive terminal. Electrons always flow anode → cathode through the wire; inside the salt bridge, anions drift toward the anode compartment and cations toward the cathode. In an electrolytic cell a power supply reverses the logic, driving a non-spontaneous reaction: the anode is now positive and the cathode negative, but the definitions never change — anode = oxidation, cathode = reduction, always (CED Topics 9.7–9.10).

Cell potentials and the standard table

Every half-reaction, written as a reduction, has a standard reduction potential E° measured against the standard hydrogen electrode at 1 M, 1 atm, 25 °C. The table is the league table of oxidising strength: the more positive the E°, the more eagerly the species is reduced (F₂, +2.87 V, is the fiercest oxidising agent on the list); the more negative, the more its reduced form wants to be oxidised (Li, E° = −3.05 V, is the strongest reducing agent). The standard cell potential is E°cell = E°(cathode) − E°(anode), using reduction potentials for both, with no sign-flipping and no multiplication by coefficients — E° is an intensive property, a voltage, not an amount. A positive E°cell means a spontaneous (galvanic) reaction; a negative value means the reaction runs the other way or needs a power supply. Two bridges follow at once: ΔG° = −nFE° (n = moles of electrons transferred per mole of reaction, F = 96,485 C/mol) and, at 25 °C, log K = nE° ÷ 0.0592. Spontaneous cells have E° > 0, ΔG° < 0 and K > 1 — the three statements are one fact in three languages.

Worked example 1: the Daniell cell, completely analysed

For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V: Identify the electrodes: zinc is oxidised (it is higher in reducing strength; its E° is more negative), so Zn is the anode and Cu is the cathode. E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = +1.10 V. Positive: the cell runs spontaneously as written. ΔG° = −nFE° = −2 × 96,485 × 1.10 = −212 kJ per mole of reaction. Equilibrium constant: log K = nE°/0.0592 = (2 × 1.10) ÷ 0.0592 = 37.2, so K ≈ 10³⁷ — the reaction goes to completion for any practical purpose, which is exactly what a battery needs. Cell notation compresses all of this into one line: Zn(s) | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu(s), anode on the left by convention.

Non-standard conditions: the Nernst equation

As a cell discharges, reactants are consumed and products accumulate, Q rises toward K, and the voltage sags. The Nernst equation prices this at 25 °C: E = E° − (0.0592 ÷ n) log Q. For the Daniell cell running with [Zn²⁺] = 1.50 M and [Cu²⁺] = 0.010 M, Q = [Zn²⁺] ÷ [Cu²⁺] = 150, so E = 1.10 − (0.0592 ÷ 2) × log(150) = 1.10 − 0.0296 × 2.18 = 1.10 − 0.064 = 1.04 V. When the cell is fully dead, Q = K and E = 0 — a flat battery is simply a reaction that has reached equilibrium. Conversely, a concentration cell (same half-reaction on both sides, different concentrations) has E° = 0 but E > 0 because Q ≠ 1: it runs until the two concentrations equalise.

Electrolysis and Faraday's law

In electrolysis the quantity of chemistry is set by the quantity of charge: Q = I × t (current in amps × time in seconds = coulombs), and each mole of electrons carries F = 96,485 C. Divide the charge by F to get moles of electrons, then use the half-reaction's ratio to get moles — and grams — of product.

Worked example 2: electroplating copper

A current of 2.50 A is passed for 30.0 minutes through aqueous CuSO₄ with inert electrodes. Calculate the mass of copper deposited at the cathode. Charge: Q = 2.50 × (30.0 × 60) = 2.50 × 1800 = 4500 C. Moles of electrons: 4500 ÷ 96,485 = 0.0466 mol. Cathode reaction: Cu²⁺ + 2 e⁻ → Cu, so moles of Cu = 0.0466 ÷ 2 = 0.0233 mol. Mass: 0.0233 × 63.55 = 1.48 g of copper. The same three-step skeleton (charge → moles of electrons → moles of product) solves every Faraday problem, including gases evolved at the other electrode and the electrolysis of molten salts, where water cannot compete for the electrons.

In the lab and in the real world

The Daniell cell is the ancestor of every battery: alkaline cells (Zn + MnO₂, 1.5 V), lead–acid car batteries (Pb + PbO₂ + H₂SO₄, 2 V per cell, six in series for 12 V — and *rechargeable* because an alternator can drive the reaction backwards, turning the galvanic cell into an electrolytic one), and lithium-ion cells, whose high voltage comes straight from lithium's position at the negative end of the E° table. Industry runs on the electrolytic side: aluminium is extracted by electrolysis of molten alumina (about 15 kWh per kilogram — which is why smelters sit next to hydroelectric dams), chlorine and sodium hydroxide come from electrolysing brine, and electroplating lays down measured thicknesses of chrome, nickel and gold priced by Faraday's law to the coulomb.

Common mistakes

• Flipping the sign of E°(anode) AND subtracting. E°cell = E°cathode − E°anode uses the table values as printed for both; students who first "reverse the anode potential" and then subtract have flipped twice and land on the wrong answer.
• Multiplying E° by the coefficients when balancing electron counts. Voltage is intensive: a half-reaction multiplied by 3 keeps the same E°. (Multiply ΔG freely — it is extensive — but never E.)
• Swapping anode and cathode signs between cell types. Oxidation is always at the anode; the *sign* of the anode is negative in a galvanic cell and positive in an electrolytic cell, because the power supply dictates the charge in the second case.
• Sending the salt-bridge ions the wrong way. Anions migrate toward the anode compartment (to balance the positive ions being produced there), cations toward the cathode compartment.
• Using seconds as minutes in Q = It, or forgetting that Cu²⁺ needs two electrons per atom. The half-reaction's electron count is part of every Faraday calculation.

AP Exam Experience

Question 1. Based on the standard reduction potentials — F₂/F⁻ +2.87 V, Cl₂/Cl⁻ +1.36 V, Cu²⁺/Cu +0.34 V, Zn²⁺/Zn −0.76 V — the strongest oxidising agent is

• (A) Zn
• (B) Cu²⁺
• (C) Cl₂
• (D) F₂

Question 2. In a working galvanic cell, which statement is always true?

• (A) The anode is positive and reduction occurs there
• (B) Oxidation occurs at the anode, which is the negative terminal
• (C) Electrons flow through the salt bridge from anode to cathode
• (D) The cathode loses mass as metal dissolves

Question 3. A current of 1.00 A flows for 965 s through a solution of AgNO₃ with inert electrodes. The mass of silver deposited is closest to (F = 96,485 C/mol; Ag = 107.9 g/mol)

• (A) 0.54 g
• (B) 1.08 g
• (C) 2.16 g
• (D) 10.8 g

Answer key. 1. (D) — the species with the most positive reduction potential is reduced most readily, i.e. it is the strongest oxidising agent. (Zn metal is the strongest *reducing* agent in the list — the mirror-image trap.) 2. (B) — oxidation is always at the anode, and in a galvanic cell the anode is negative; electrons travel through the external wire, not the salt bridge (ions move there), and a cathode typically *gains* mass as metal plates out. 3. (B) — Q = 1.00 × 965 = 965 C; moles of electrons = 965 ÷ 96,485 = 0.0100 mol; Ag⁺ + e⁻ → Ag is 1 : 1, so 0.0100 mol × 107.9 = 1.08 g.

Free-response question. A galvanic cell is constructed as Cd(s) | Cd²⁺(1.0 M) || Cu²⁺(1.0 M) | Cu(s). Standard reduction potentials: Cu²⁺/Cu +0.34 V; Cd²⁺/Cd −0.40 V. (a) Write the half-reaction at each electrode and the overall cell reaction, labelling anode and cathode. (b) Calculate E°cell. (c) Calculate ΔG° for the overall reaction. (d) The cell runs until [Cu²⁺] has fallen to 0.010 M while [Cd²⁺] remains approximately 1.0 M. Calculate the cell potential under these conditions. (e) During the run in part (d), exactly 1.00 mol of electrons passes through the external circuit. Calculate the change in mass of the cadmium electrode. (Cd = 112.4 g/mol.) Scoring guideline.

• (a) 1 point for anode (oxidation): Cd(s) → Cd²⁺(aq) + 2 e⁻; cathode (reduction): Cu²⁺(aq) + 2 e⁻ → Cu(s); overall: Cd(s) + Cu²⁺(aq) → Cd²⁺(aq) + Cu(s).
• (b) 1 point for E°cell = 0.34 − (−0.40) = +0.74 V.
• (c) 1 point for ΔG° = −nFE° = −2 × 96,485 × 0.74 = −1.43 × 10⁵ J = −143 kJ, negative as required for a galvanic cell.
• (d) 1 point for the Nernst equation with Q = [Cd²⁺]/[Cu²⁺] = 1.0/0.010 = 100: E = 0.74 − (0.0592/2) × log(100) = 0.74 − 0.0296 × 2 = 0.68 V.
• (e) 1 point for moles of Cd dissolved = 1.00 mol e⁻ ÷ 2 = 0.500 mol, mass = 0.500 × 112.4 = 56.2 g, so the cadmium electrode loses 56.2 g (it is consumed at the anode).