Electrochemistry: electrolysis, redox processes, standard electrode potentials, fuel cells

Electrochemistry

Learning Objectives

• Explain the principles of electrolysis and the factors that influence product distribution.
• Write, balance and analyse redox (oxidation–reduction) reactions using oxidation numbers in both acidic and basic media.
• Describe galvanic (voltaic) and electrolytic cells, use proper cell notation and calculate the standard electromotive force (E°_cell).
• Interpret standard electrode potentials, predict the spontaneity of redox processes and relate E°_cell to ΔG°.
• Derive and apply the Nernst equation for non‑standard conditions.
• Explain the operation of common fuel cells and calculate their theoretical cell potentials.

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1. Key Conventions (Box)

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2. Electrochemical Cells

2.1 Galvanic (Voltaic) vs. Electrolytic Cells

Feature	Galvanic (Spontaneous)	Electrolytic (Non‑spontaneous)
Energy flow	Chemical → Electrical	Electrical → Chemical
External voltage required	No	Yes (to overcome ΔG° > 0)
Anode sign	Negative	Positive
Cathode sign	Positive	Negative
Typical use	Batteries, corrosion	Electroplating, electro‑refining, production of gases

2.2 Cell Notation

General form (left = anode, right = cathode):

metal | metal‑ion (aq, c) || metal‑ion (aq, c) | metal

Example for a Zn/Cu cell:

Zn(s) | Zn²⁺(1 M) ‖ Cu²⁺(1 M) | Cu(s)

2.3 Calculating the Standard EMF (E°_cell)

E°_cell = E°_cathode – E°_anode

Both potentials are taken from the standard reduction‑potential table (relative to SHE).

2.4 Worked Example – Zn/Cu Cell

Half‑cell (reduction)	E° (V)
Zn²⁺ + 2 e⁻ → Zn(s)	–0.76
Cu²⁺ + 2 e⁻ → Cu(s)	+0.34

Cell notation: Zn(s) | Zn²⁺(1 M) ‖ Cu²⁺(1 M) | Cu(s)

Calculation:

• E°_cathode = +0.34 V (Cu²⁺/Cu)
• E°_anode = –0.76 V (Zn²⁺/Zn)
• E°_cell = 0.34 – (–0.76) = **+1.10 V** → spontaneous.

Overall reaction:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

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3. Redox Processes

3.1 Oxidation‑Number Rules (quick reference)

Rule	Oxidation number
Elements in their standard state	0
Monatomic ions	equal to the ionic charge
Oxygen	–2 (except in peroxides –1, OF₂ +2)
Hydrogen	+1 (except metal hydrides –1)
Fluorine	–1 (always)
Other halogens	–1 unless bonded to O or another halogen of higher electronegativity
Sum of oxidation numbers = overall charge of the species	–

3.2 Balancing Redox Equations – Half‑Reaction Method

3.2.1 In Acidic Medium

1. Assign oxidation numbers and split the overall reaction into oxidation and reduction half‑reactions.
2. Balance all atoms **except** H and O.
3. Balance O by adding H₂O.
4. Balance H by adding H⁺.
5. Balance charge by adding electrons (e⁻).
6. Multiply half‑reactions to equalise the number of electrons transferred.
7. Add the half‑reactions and cancel species that appear on both sides.

3.2.2 In Basic Medium

1. Carry out steps 1–6 exactly as for acidic medium.
2. When H⁺ appears in the balanced equation, neutralise each H⁺ by adding an equal number of OH⁻ to both sides.
3. Combine H⁺ + OH⁻ → H₂O on the side where they meet, then cancel any water molecules that appear on both sides.
4. Check that O and H are balanced and that the charge is balanced.

Example – Balancing in Basic Medium

Reaction: MnO₄⁻ + H₂O → MnO₂ + OH⁻

1. Identify oxidation states: Mn +7 → +4 (reduction).
2. Half‑reaction (reduction):
     MnO₄⁻ → MnO₂
3. Balance O by adding H₂O:
     MnO₄⁻ → MnO₂ + 2 H₂O
4. Balance H by adding OH⁻ (basic medium):
     MnO₄⁻ + 4 OH⁻ → MnO₂ + 2 H₂O + 4 OH⁻ → cancel 2 H₂O on each side →
     MnO₄⁻ + 2 OH⁻ → MnO₂ + 2 OH⁻
5. Balance charge by adding electrons:
     MnO₄⁻ + 2 OH⁻ + 3 e⁻ → MnO₂ + 2 OH⁻
6. Final balanced half‑reaction (simplified):
     MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻

3.3 Example – Permanganate Oxidises Ferrous Ion (Acidic)

Half‑reactions

• Reduction: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
• Oxidation: Fe²⁺ → Fe³⁺ + e⁻

Multiply oxidation by 5 and add:

2 MnO₄⁻ + 10 Fe²⁺ + 16 H⁺ → 2 Mn²⁺ + 10 Fe³⁺ + 8 H₂O

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4. Standard Electrode Potentials

All values are measured under standard conditions (1 M, 1 atm, 25 °C) and are expressed relative to the standard hydrogen electrode (SHE, 0.00 V).

Half‑cell (reduction)	E° (V)
2 H⁺ + 2 e⁻ → H₂                 (SHE)	0.00
Zn²⁺ + 2 e⁻ → Zn(s)	–0.76
Cu²⁺ + 2 e⁻ → Cu(s)	+0.34
Ag⁺ + e⁻ → Ag(s)	+0.80
Fe³⁺ + e⁻ → Fe²⁺	+0.77
Fe²⁺ + 2 e⁻ → Fe(s)	–0.44
Cl₂ + 2 e⁻ → 2 Cl⁻	+1.36
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O	+1.51
Pb²⁺ + 2 e⁻ → Pb(s)	–0.13
Ni²⁺ + 2 e⁻ → Ni(s)	–0.25

4.1 Predicting Spontaneity

• Galvanic cell: E°_cell > 0 → ΔG° < 0 → spontaneous.
• Electrolytic cell: E°_cell < 0 → ΔG° > 0 → non‑spontaneous; external voltage must be applied.

Relationship:

ΔG° = –n F E°_cell

where n = electrons transferred, F = 96 485 C mol⁻¹.

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5. The Nernst Equation

For non‑standard conditions:

E = E° – (RT / nF) ln Q

• R = 8.314 J mol⁻¹ K⁻¹
• T = temperature in kelvin
• n = electrons transferred in the overall cell reaction
• F = 96 485 C mol⁻¹
• Q = reaction quotient (activities of products ÷ activities of reactants)

At 25 °C (298 K) the base‑10 form is often used:

E = E° – (0.0592 V / n) log₁₀ Q

Example – Nernst Calculation for the Zn/Cu Cell

Given: [Zn²⁺] = 0.010 M, [Cu²⁺] = 0.10 M, T = 298 K.

Overall reaction: Zn + Cu²⁺ → Zn²⁺ + Cu

Q = [Zn²⁺] / [Cu²⁺] = 0.010 / 0.10 = 0.10

E = 1.10 V – (0.0592 V / 2) log₁₀(0.10) = 1.10 V – 0.0296 V (–1) = **1.13 V**

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6. Electrolysis

6.1 Fundamental Concepts

1. Anode (positive) – oxidation occurs.
2. Cathode (negative) – reduction occurs.
3. Electrolyte – provides mobile ions; conductivity depends on concentration and ion mobility.
4. Over‑potential – extra voltage required to initiate gas evolution or to overcome activation barriers.

6.2 Typical Products from Common Electrolytes

Electrolyte	Anode (oxidation)	Cathode (reduction)	Overall reaction
Molten NaCl	2 Cl⁻ → Cl₂(g) + 2 e⁻	Na⁺ + e⁻ → Na(l)	2 NaCl(l) → Cl₂(g) + 2 Na(l)
Aqueous NaCl	2 Cl⁻ → Cl₂(g) + 2 e⁻	2 H₂O + 2 e⁻ → H₂(g) + 2 OH⁻	2 NaCl(aq) + 2 H₂O(l) → Cl₂(g) + H₂(g) + 2 Na⁺(aq) + 2 OH⁻(aq)
Aqueous CuSO₄	2 H₂O → O₂(g) + 4 H⁺ + 4 e⁻	Cu²⁺ + 2 e⁻ → Cu(s)	2 Cu²⁺(aq) + 2 H₂O(l) → 2 Cu(s) + O₂(g) + 4 H⁺(aq)

6.3 Factors Influencing Product Distribution

• Standard electrode potentials – the species with the more positive reduction potential is reduced preferentially at the cathode.
• Concentration of ions – higher ion concentration shifts the Nernst potential, possibly changing which half‑reaction dominates.
• Over‑potential – can suppress the evolution of a gas with a higher thermodynamic potential (e.g., chlorine vs. oxygen).
• Nature of the electrolyte (aqueous vs. molten) – determines whether water or the anion is oxidised at the anode.

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7. Fuel Cells

7.1 Hydrogen–Oxygen (PEM) Fuel Cell

Overall reaction (standard conditions)

2 H₂(g) + O₂(g) → 2 H₂O(l)

Half‑reactions

• Anode (oxidation): H₂(g) → 2 H⁺ + 2 e⁻  E°_anode = 0.00 V (SHE)
• Cathode (reduction): ½ O₂(g) + 2 H⁺ + 2 e⁻ → H₂O(l)  E°_cathode = +1.23 V

E°_cell = 1.23 V

ΔG° = –n F E°_cell = –2 × 96 485 C mol⁻¹ × 1.23 V ≈ –2.37 × 10⁵ J mol⁻¹

7.2 Other Commercial Fuel‑Cell Types (A‑Level focus)

1. Proton Exchange Membrane (PEM) Fuel Cell – solid polymer electrolyte; operates at 60–80 °C; fast H₂ oxidation kinetics.
2. Alkaline Fuel Cell (AFC) – aqueous KOH electrolyte; high theoretical efficiency; sensitive to CO₂ (forms carbonate).
3. Solid Oxide Fuel Cell (SOFC) – ceramic oxide electrolyte (e.g., YSZ); operates at 800–1000 °C; can use hydrocarbons after reforming.

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8. Summary Checklist

• Identify anode (oxidation) and cathode (reduction) for any electrochemical process; write the corresponding half‑reactions.
• Use correct cell notation and calculate E°_cell from standard reduction potentials.
• Balance redox equations by the half‑reaction method in both acidic and basic media.
• Apply ΔG° = –n F E°_cell to decide whether a reaction is spontaneous.
• Use the Nernst equation to find cell potentials when concentrations, pressures or temperature differ from standard conditions.
• Explain over‑potential and other factors that affect product distribution in electrolysis.
• Describe the operation, overall reaction and theoretical voltage of a hydrogen fuel cell, and name the main commercial fuel‑cell types.