Electrochemistry

Oxidation & Reduction Recap

Electrochemistry is built on redox reactions. Remember OIL RIG: Oxidation Is Loss of electrons; Reduction Is Gain of electrons. In electrochemistry, electron transfer is harnessed to do electrical work (galvanic cell) or driven by electrical work (electrolytic cell).

Anode = Oxidation (A–O). Cathode = Reduction (C–R). Both start with the same letter pair: A–O and C–R. This holds for BOTH galvanic and electrolytic cells.

Galvanic (Voltaic) Cells

A galvanic cell converts the chemical energy of a spontaneous redox reaction into electrical energy. The Daniell cell (Zn–Cu) is the classic example:

Anode (negative electrode): Zinc electrode in ZnSO₄ solution. Oxidation: Zn → Zn²⁺ + 2e⁻. Zn dissolves; electrode gets lighter.
Cathode (positive electrode): Copper electrode in CuSO₄ solution. Reduction: Cu²⁺ + 2e⁻ → Cu. Cu deposits; electrode gets heavier.
External circuit: Electrons flow from anode (Zn) through the wire to cathode (Cu).
Salt bridge: Contains KNO₃ or KCl in agar. Allows ion migration to maintain electrical neutrality in both half-cells. Without it, the cell stops working. Cations migrate toward cathode half-cell; anions migrate toward anode half-cell.

Anode: Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Cell notation (line notation): Anode on the left; cathode on the right; double vertical line (||) represents the salt bridge; single vertical line (|) represents a phase boundary.

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Standard Electrode Potentials (E°)

The standard electrode potential (E°) is the potential of a half-cell measured against the standard hydrogen electrode (SHE, E° = 0.00 V) under standard conditions: 25°C, 1 mol·dm⁻³, 1 atm. All values are written as reduction potentials.

Half-reaction (reduction)	E° (V)	Character
F₂ + 2e⁻ → 2F⁻	+2.87	Strongest OA
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	+1.51	Strong OA
Cl₂ + 2e⁻ → 2Cl⁻	+1.36	Strong OA
O₂ + 4H⁺ + 4e⁻ → 2H₂O	+1.23
Ag⁺ + e⁻ → Ag	+0.80
Fe³⁺ + e⁻ → Fe²⁺	+0.77
Cu²⁺ + 2e⁻ → Cu	+0.34
2H⁺ + 2e⁻ → H₂	0.00	SHE (reference)
Fe²⁺ + 2e⁻ → Fe	−0.44
Zn²⁺ + 2e⁻ → Zn	−0.76	Strong RA
Na⁺ + e⁻ → Na	−2.71	Strong RA
Li⁺ + e⁻ → Li	−3.04	Strongest RA

E°_cell = E°_cathode − E°_anode

Rule: If E°_cell > 0, the reaction is spontaneous under standard conditions (galvanic cell). If E°_cell < 0, the reaction is non-spontaneous (requires electrical input → electrolytic cell). For the Zn-Cu cell: E°_cell = +0.34 − (−0.76) = +1.10 V.

Electrolytic Cells

An electrolytic cell uses an external power supply (battery/DC source) to drive a non-spontaneous redox reaction. The positive terminal of the battery is connected to the anode; the negative terminal to the cathode.

Anode: Oxidation occurs (connected to + terminal of power supply).
Cathode: Reduction occurs (connected to − terminal of power supply).

Electrolysis of molten NaCl (Chlor-alkali process):

Cathode (−): Na⁺ + e⁻ → Na (sodium metal)
Anode (+): 2Cl⁻ → Cl₂ + 2e⁻ (chlorine gas)

Electrolysis of water (dilute H₂SO₄ as electrolyte):

Cathode: 4H⁺ + 4e⁻ → 2H₂(g) (hydrogen gas, 2 volumes)
Anode: 2H₂O → O₂(g) + 4H⁺ + 4e⁻ (oxygen gas, 1 volume)

Electroplating: A metal object (cathode) is coated with another metal. The plating metal is the anode. Example — silver plating a spoon: silver anode, spoon as cathode, silver nitrate solution as electrolyte. Ag⁺ ions deposit on the spoon at the cathode: Ag⁺ + e⁻ → Ag.

Faraday’s Laws of Electrolysis

Faraday’s First Law: The mass of substance deposited at an electrode is proportional to the quantity of charge passed.

Faraday’s Second Law: For the same quantity of charge, the mass deposited is proportional to the molar mass and inversely proportional to the number of electrons transferred (n).

Q = It      (charge = current × time, in coulombs)
n_substance = Q / (n_e × F)      (moles deposited)
F = 96 485 C·mol⁻¹      (Faraday constant)

Worked example: How many grams of copper are deposited when a current of 2.00 A flows for 30 minutes through a CuSO₄ solution?

Q = It = 2.00 × (30 × 60) = 3 600 C
Cu²⁺ + 2e⁻ → Cu (n_e = 2)
n(Cu) = Q / (n_e × F) = 3600 / (2 × 96485) = 0.01865 mol
m(Cu) = 0.01865 × 63.5 = 1.18 g

⭐ IEB Extension — Nernst Equation & Concentration Cells

The standard cell potential E° applies only under standard conditions. The Nernst equation calculates the actual cell potential at non-standard concentrations:

E = E° − (RT/nF) ln Q ≈ E° − (0.0592/n) log Q (at 25°C)

Where R = 8.314 J·mol⁻¹·K⁻¹, T = temperature (K), n = moles of electrons transferred, F = 96 485 C·mol⁻¹, Q = reaction quotient.

If Q < 1 (products low, reactants high): E > E° — cell potential is boosted.
If Q > 1 (products high): E < E° — cell potential is reduced.
At equilibrium: E = 0 and Q = K. Battery is “flat”.

Concentration cells: Both half-cells use the same electrode and ion, but at different concentrations. E°_cell = 0, but E_cell ≠ 0 because the Nernst equation gives a non-zero value. Current flows from the dilute to the concentrated half-cell until concentrations equalise.

Answer all questions in your notebook. Show full calculations with units. For cell diagrams, label all parts: anode, cathode, direction of electron flow, salt bridge, and ion migration directions.

Question 1 — Galvanic Cell Analysis

A galvanic cell is constructed using Fe and Ag electrodes in their respective ion solutions.
Given: E°(Ag⁺/Ag) = +0.80 V; E°(Fe²⁺/Fe) = −0.44 V

(a) Identify the anode and cathode. Justify your answer using E° values.
(b) Write the half-reaction at each electrode and the overall balanced cell reaction.
(c) Calculate E°_cell. Is the reaction spontaneous? Explain.
(d) Write the cell notation (line notation) for this cell.
(e) Describe the direction of electron flow and ion migration through the salt bridge.

(10 marks)

Question 2 — Electrolysis Calculations (Faraday)

A current of 3.00 A is passed through a solution of silver nitrate (AgNO₃) for 45 minutes.
Given: Ag⁺ + e⁻ → Ag; M(Ag) = 108 g·mol⁻¹; F = 96 485 C·mol⁻¹

(a) Calculate the total charge Q passed.
(b) Calculate the moles of Ag deposited.
(c) Calculate the mass of Ag deposited.
(d) At which electrode (anode or cathode) does the silver deposit? Write the half-reaction.

(6 marks)

Question 3 — Predicting Reactions Using E°

Use the table of standard electrode potentials to answer the following:

(a) Can Cl₂ oxidise Fe²⁺ to Fe³⁺? Show your calculation of E°_cell.
(b) Will zinc dissolve in a solution of CuSO₄? Write the cell reaction and calculate E°_cell.
(c) Explain why copper does not dissolve in dilute HCl but zinc does. Use E° values.

(6 marks)

Question 4 — Electroplating

A jeweller wants to electroplate a silver bracelet with gold (Au). The electrolyte used is gold(III) chloride solution (AuCl₃).

(a) Identify the anode, cathode, and electrolyte in this electroplating cell.
(b) Write the half-reaction that occurs at the cathode.
(c) How much charge would be needed to deposit 0.500 g of gold? (M(Au) = 197 g·mol⁻¹; Au³⁺ + 3e⁻ → Au)
(d) If a current of 0.250 A is used, how long (in minutes) would the plating take?

(6 marks)

Question 5 — Reading Electroplating Data

A learner electroplates copper from a CuSO₄ solution (Cu²⁺ + 2e⁻ → Cu; M(Cu) = 64 g·mol⁻¹) for a fixed time of 25 minutes, using different currents:

Current I (A)	0.5	1.0	1.5	2.0
Mass of Cu deposited (g)	0.25	0.50	0.75	1.00

(a) Calculate mass ÷ I for each column. What does this (approximately constant) value confirm about the relationship between mass deposited and current, at fixed time?
(b) Using F = 96 485 C·mol⁻¹ and the I = 1.0 A data point, calculate n(Cu) deposited, then n(e⁻) passed, then the charge Q. Show that this charge is consistent with a plating time of about 25 minutes at 1.0 A.
(c) Using the pattern in the table (not a fresh calculation from F), predict the mass of copper that would be deposited by a current of 3.0 A over the same 25 minutes.

(6 marks)

Oxidation & Reduction Recap

Galvanic (Voltaic) Cells

Standard Electrode Potentials (E°)

Electrolytic Cells

Faraday’s Laws of Electrolysis

Daniell Cell (Galvanic) — Zn | Zn²♠ || Cu²♠ | Cu