A Level Chemistry - How to Solve: Electrode Potentials & Electrochemical Cells (Nernst Equation, Cell EMF, Fuel Cells)

How to Solve: Electrode Potentials & Electrochemical Cells (Nernst Equation, Cell EMF, Fuel Cells)

Complete Guide for GCSE/A-Level (Physics, Chemistry, Biology)

Introduction

"Mastering electrode potentials and the Nernst equation lets you predict whether a battery will work, design fuel cells for electric cars, and answer 10–15% of your A-Level Chemistry exam—worth up to 18 marks. Get this right, and you’re one step closer to an A."

WHAT YOU NEED TO KNOW FIRST

Before diving in, you must understand:
1. Redox reactions – Oxidation (loss of electrons) and reduction (gain of electrons).
2. Standard electrode potentials (E°) – How to read and interpret a table of reduction potentials.
3. Basic cell notation – How to write and interpret electrochemical cell diagrams (e.g., Zn | Zn²⁺ || Cu²⁺ | Cu).

If you’re shaky on any of these, pause and review them now—this guide won’t make sense without them.

KEY TERMS & FORMULAS

Key Terms

Formulas

1. Cell EMF (Standard Conditions) E°_cell = E°_{reduction (cathode)} – E°_{reduction (anode)}
2. E°_{reduction (cathode)} = More positive E° value (where reduction happens).
3. E°_{reduction (anode)} = More negative E° value (where oxidation happens).

4. MEMORISE THIS – This is the most important formula for cell EMF.

5. Nernst Equation (Non-Standard Conditions) E = E° – (RT / nF) × ln(Q)

6. E = Electrode potential under non-standard conditions (V).
7. E° = Standard electrode potential (V).
8. R = Gas constant (8.314 J/mol·K) – given on exam sheet.
9. T = Temperature in Kelvin (K).
10. n = Number of electrons transferred in the reaction.
11. F = Faraday’s constant (96,485 C/mol) – given on exam sheet.
12. Q = Reaction quotient (concentration of products / reactants, raised to their stoichiometric coefficients).

13. Simplified version (at 298 K): E = E° – (0.059 / n) × log(Q)

    • MEMORISE THIS – Examiners love this simplified form.

14. Gibbs Free Energy & Cell EMF ΔG° = –nFE°_cell

15. ΔG° = Standard Gibbs free energy change (J/mol).
16. n = Number of electrons transferred.
17. F = Faraday’s constant (96,485 C/mol).
18. E°_cell = Standard cell EMF (V).
19. MEMORISE THIS – Links thermodynamics to electrochemistry.

STEP-BY-STEP METHOD

How to Calculate Cell EMF (E°_cell)

Step 1: Write the half-equations for the two half-cells. Step 2: Identify the cathode (where reduction happens) and anode (where oxidation happens). - Cathode = More positive E° value. - Anode = More negative E° value. Step 3: Write the overall cell reaction by balancing electrons. Step 4: Calculate E°_cell = E°_cathode – E°_anode. Step 5: If E°_cell > 0, the reaction is spontaneous.

How to Use the Nernst Equation

Step 1: Write the balanced half-reaction for the electrode. Step 2: Identify n (number of electrons transferred). Step 3: Write Q (reaction quotient) using concentrations (for gases, use partial pressures). - For Zn²⁺ + 2e⁻ → Zn, Q = 1 / [Zn²⁺] (solids are omitted). - For 2H⁺ + 2e⁻ → H₂, Q = P_H₂ / [H⁺]². Step 4: Plug into the Nernst equation: E = E° – (0.059 / n) × log(Q) (at 298 K). Step 5: Calculate E and interpret: - If E > E°, the reaction is more favourable under these conditions. - If E < E°, the reaction is less favourable.

How to Analyse Fuel Cells

Step 1: Write the half-reactions for the anode and cathode. - Anode (oxidation): H₂ → 2H⁺ + 2e⁻ (in a hydrogen fuel cell). - Cathode (reduction): O₂ + 4H⁺ + 4e⁻ → 2H₂O. Step 2: Balance the overall reaction (ensure electrons cancel). Step 3: Calculate E°_cell using standard potentials. Step 4: Use the Nernst equation if conditions are non-standard (e.g., different pressures). Step 5: State advantages/disadvantages of fuel cells (e.g., high efficiency, but expensive catalysts).

WORKED EXAMPLES

Example 1 – Basic: Calculating Cell EMF

Question: Calculate the standard cell EMF for the following cell: Zn | Zn²⁺ (1 mol/dm³) || Cu²⁺ (1 mol/dm³) | Cu Given: - E°(Zn²⁺/Zn) = –0.76 V - E°(Cu²⁺/Cu) = +0.34 V

Solution: Step 1: Write half-equations. - Zn → Zn²⁺ + 2e⁻ (oxidation, anode) - Cu²⁺ + 2e⁻ → Cu (reduction, cathode)

Step 2: Identify cathode and anode. - Cathode (reduction): Cu²⁺/Cu (E° = +0.34 V) - Anode (oxidation): Zn²⁺/Zn (E° = –0.76 V)

Step 3: Calculate E°_cell. E°_cell = E°_cathode – E°_anode = +0.34 V – (–0.76 V) = +1.10 V

Step 4: Interpret. - E°_cell > 0, so the reaction is spontaneous.

What we did and why: We used the standard cell EMF formula to find the voltage of a zinc-copper cell. The more positive E° value (Cu) is the cathode, and the more negative (Zn) is the anode. The positive EMF confirms the reaction happens naturally.

Example 2 – Medium: Nernst Equation (Non-Standard Conditions)

Question: Calculate the electrode potential for the Zn²⁺/Zn half-cell when [Zn²⁺] = 0.1 mol/dm³ at 298 K. Given: E°(Zn²⁺/Zn) = –0.76 V

Solution: Step 1: Write the half-reaction. Zn²⁺ + 2e⁻ → Zn

Step 2: Identify n (number of electrons). n = 2

Step 3: Write Q (reaction quotient). - Solids (Zn) are omitted. - Q = 1 / [Zn²⁺] = 1 / 0.1 = 10

Step 4: Plug into the Nernst equation (simplified, 298 K). E = E° – (0.059 / n) × log(Q) = –0.76 V – (0.059 / 2) × log(10) = –0.76 V – (0.0295) × 1 = –0.7895 V

Step 5: Interpret. - The potential is more negative than E° because the lower [Zn²⁺] makes oxidation less favourable.

What we did and why: We used the Nernst equation to adjust the standard potential for a non-standard concentration. Lower [Zn²⁺] shifts the equilibrium left, making the electrode potential more negative.

Example 3 – Exam-Style: Fuel Cell EMF

Question: A hydrogen-oxygen fuel cell operates at 298 K with P_H₂ = 0.5 atm and [H⁺] = 0.1 mol/dm³. Calculate the cell EMF under these conditions. Given: - E°(O₂/H₂O) = +1.23 V - E°(H⁺/H₂) = 0 V

Solution: Step 1: Write half-reactions. - Anode (oxidation): H₂ → 2H⁺ + 2e⁻ - Cathode (reduction): O₂ + 4H⁺ + 4e⁻ → 2H₂O

Step 2: Balance electrons (multiply anode by 2). - 2H₂ → 4H⁺ + 4e⁻ - O₂ + 4H⁺ + 4e⁻ → 2H₂O

Step 3: Calculate E°_cell. E°_cell = E°_cathode – E°_anode = +1.23 V – 0 V = +1.23 V

Step 4: Write Q for the overall reaction. Overall: 2H₂ + O₂ → 2H₂O - Q = 1 / (P_H₂² × P_O₂ × [H⁺]⁴) - Assume P_O₂ = 1 atm (not given, so standard). - Q = 1 / (0.5² × 1 × 0.1⁴) = 1 / (0.25 × 1 × 0.0001) = 40,000

Step 5: Use the Nernst equation (n = 4). E = E° – (0.059 / n) × log(Q) = 1.23 V – (0.059 / 4) × log(40,000) = 1.23 V – (0.01475) × 4.602 = 1.23 V – 0.0679 V = 1.162 V

Step 6: Interpret. - The lower EMF (1.162 V vs. 1.23 V) is due to non-standard conditions (lower P_H₂ and [H⁺]).

What we did and why: We combined cell EMF and the Nernst equation to adjust for non-standard pressures and concentrations in a fuel cell. The lower EMF shows how real-world conditions reduce efficiency.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before the Exam)

"Right, listen up—this is your electrode potentials cheat sheet in 60 seconds. First, cell EMF: cathode minus anode, always. More positive E° = cathode. If E°_cell is positive, it’s spontaneous. Second, Nernst equation: E = E° – (0.059/n) × log(Q). Q is products over reactants, no solids. Third, fuel cells: H₂ oxidised at anode, O₂ reduced at cathode. Non-standard conditions? Nernst again. Finally, Gibbs free energy: ΔG° = –nFE°_cell. Positive EMF = negative ΔG = spontaneous. Got it? Now go smash that exam."