Redox Reactions & Electrochemistry
Redox reactions involve the transfer of electrons between chemical species. In electrochemistry, oxidation and reduction half-reactions are separated into two half-cells connected by a wire and salt bridge, allowing us to measure standard electrode potentials against the standard hydrogen electrode and predict reaction feasibility.
- Oxidizing Agent
- A species that accepts electrons and gets reduced in a chemical reaction.
- Reducing Agent
- A species that donates electrons and gets oxidized in a chemical reaction.
- Standard Electrode Potential (E°)
- The voltage measured when a half-cell is connected to a standard hydrogen electrode under standard conditions.
Electrochemical Cell Builder & E°cell Calculator
Select any two half-cells below to calculate standard cell potential E°cell = E°(reduced) − E°(oxidised) and watch which way electrons flow.
Standard conditions: 298 K (25°C), 100 kPa pressure, 1.00 mol dm⁻³ ion concentrations.
E°cell = (+0.34) − (-0.76) = +1.10 V
- Anode (Oxidation: more negative E° = -0.76 V):
Zn loses electrons and dissolves into Zn²⁺(aq). Electrons flow out of the Zn electrode into the external wire.
+ Cathode (Reduction, more positive E° = +0.34 V):
Cu²⁺(aq) ions gain electrons at the surface to form solid Cu.
Exam notes: Electrode Potentials, Feasibility & Fuel Cells
- Always write half-equations as reduction processes (electrons on the left: Ox + ne⁻ ⇌ Red). The half-cell with the more negative E° value goes backwards (is oxidised, loses electrons, acts as the negative anode). The half-cell with the more positive E° value goes forwards (is reduced, gains electrons, acts as the positive cathode).
- Standard Electrode Potentials are measured against the Standard Hydrogen Electrode (SHE: E° = 0.00 V exactly) at 298 K, 100 kPa H₂(g), and 1.00 mol dm⁻³ H⁺(aq) using an inert platinum (Pt) electrode coated in platinum black.
- Applying Le Chatelier's principle to a half-cell (Ox + ne⁻ ⇌ Red): increasing the concentration of the oxidised species [Ox] shifts equilibrium to the right, making E more positive. Increasing temperature or decreasing [Ox] shifts it left, making E more negative.
| Cell Type | How It Works | Key Exam Pros & Cons |
|---|---|---|
| Non-rechargeable (e.g. Alkaline Zn/MnO₂) | Irreversible redox reactions; stops working once reactants are used up | Cheap and portable, but creates landfill waste when flat |
| Rechargeable (e.g. Lithium-ion, Lead-acid) | External electrical current reverses the anode and cathode redox reactions | Reusable hundreds of times; Li-ion is very light (low density of Li) with high voltage, but can overheat |
| Hydrogen-Oxygen Fuel Cell | Continuous external supply of H₂ fuel and O₂; only product is water (2H₂ + O₂ → 2H₂O) | Zero CO₂ emissions at point of use and doesn't need recharging, but H₂ is flammable/hard to store and often made from fossil fuels |
Key Concepts & Exam Definitions (8)
- Oxidizing Agent
- A species that accepts electrons and gets reduced in a chemical reaction.
- Reducing Agent
- A species that donates electrons and gets oxidized in a chemical reaction.
- Standard Electrode Potential (E°)
- The voltage measured when a half-cell is connected to a standard hydrogen electrode under standard conditions.
- Salt Bridge
- A connection between two half-cells, usually filter paper soaked in KNO3, that allows ions to flow and complete the circuit.
- Electrochemical Cell
- A device capable of either generating electrical energy from chemical reactions or using electrical energy to cause chemical reactions.
- Hydrogen-oxygen fuel cell half-equations (acidic vs alkaline)
- In acidic conditions: Anode (−): H₂ → 2H⁺ + 2e⁻; Cathode (+): O₂ + 4H⁺ + 4e⁻ → 2H₂O. In alkaline conditions: Anode (−): H₂ + 2OH⁻ → 2H₂O + 2e⁻; Cathode (+): O₂ + 2H₂O + 4e⁻ → 4OH⁻. Overall equation in both cases: 2H₂ + O₂ → 2H₂O (E°cell = +1.23 V).
- Conventional Cell Representation (IUPAC Notation)
- Written as: Anode (oxidation) on the left || Cathode (reduction) on the right. Double vertical line || represents the salt bridge; single vertical line | represents a phase boundary. Example for Zn/Cu cell: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).
- Why a Feasibility Prediction (E°cell > 0) Might Not Happen in Practice
- 1) The reaction may have a very high activation energy (E_a), so the rate is kinetically too slow at room temperature. 2) The actual conditions may be non-standard (concentrations not 1.00 mol dm⁻³, shifting the electrode potentials).
Quick-Fire Flashcards (9)
Click any card to check the answer, then mark it as Known to track your progress.
What are the standard conditions for measuring electrode potentials?
If a half-cell has a more negative E° value, what does it mean?
Why is a platinum electrode used in half-cells without a solid metal?
What direction do electrons flow in the external circuit of an electrochemical cell?
What is the oxidation state of oxygen in a peroxide (e.g., H2O2)?
Write the conventional cell representation for a Daniell cell made of Zn²⁺/Zn (E° = −0.76 V) and Cu²⁺/Cu (E° = +0.34 V).
Give two reasons why a redox reaction with a positive E°cell (+0.45 V) might not visibly occur in a test tube.
Why is a platinum electrode used in a Fe³⁺(aq)/Fe²⁺(aq) half-cell or the Standard Hydrogen Electrode?
What is the overall reaction and only chemical product of a hydrogen-oxygen fuel cell?
Exam-Style Questions & Mark Schemes (5)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Explain how a salt bridge maintains electrical neutrality in an electrochemical cell.
Click “Check Mark Scheme” to compare your answer, As electrons move from the negative half-cell (where oxidation occurs) to the positive half-cell (where reduction occurs), a charge imbalance builds up. Positive ions are produced at the negative electrode, and positive ions are consumed at the positive electrode. The salt bridge allows negative ions (e.g., NO3-) to flow to the negative half-cell and positive ions (e.g., K+) to flow to the positive half-cell, balancing the charges and completing the circuit.
- Q2
Predict whether zinc metal will react with aqueous copper(II) ions given: Zn2+/Zn E° = -0.76 V; Cu2+/Cu E° = +0.34 V.
Click “Check Mark Scheme” to compare your answer, Yes, they will react. Zn has a more negative E° value, so it is a stronger reducing agent and will be oxidized to Zn2+, releasing electrons. Cu2+ has a more positive E° value, so it will be reduced to Cu by accepting those electrons. The overall cell potential is +0.34 - (-0.76) = +1.10 V. A positive E°cell indicates a feasible reaction.
- Q3
Balance the redox equation for the reaction between acidified manganate(VII) ions (MnO4-) and iron(II) ions (Fe2+).
Click “Check Mark Scheme” to compare your answer, Reduction half-equation: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O. Oxidation half-equation: Fe2+ -> Fe3+ + e-. Multiply the oxidation equation by 5 to balance electrons: 5Fe2+ -> 5Fe3+ + 5e-. Combine the equations: MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+.
- Q4
Given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Cl₂/Cl⁻) = +1.36 V, state with a reason whether chlorine can oxidise Fe²⁺(aq) ions to Fe³⁺(aq), and write the overall balanced equation. [3 marks]
Click “Check Mark Scheme” to compare your answer, 1) E°(Cl₂/Cl⁻) (+1.36 V) is more positive than E°(Fe³⁺/Fe²⁺) (+0.77 V), so E°cell = 1.36 − 0.77 = +0.59 V (> 0, feasible) [1 mark]. 2) Cl₂ is reduced to 2Cl⁻ and Fe²⁺ is oxidised to Fe³⁺ [1 mark]. 3) Overall balanced equation: Cl₂(g) + 2Fe²⁺(aq) → 2Cl⁻(aq) + 2Fe³⁺(aq) [1 mark].
- Q5
In a Zn / Cu electrochemical cell, predict and explain how the cell EMF changes if water is added to the Zn²⁺(aq) half-cell (diluting [Zn²⁺]). [3 marks]
Click “Check Mark Scheme” to compare your answer, 1) For the equilibrium Zn²⁺(aq) + 2e⁻ ⇌ Zn(s), decreasing [Zn²⁺] shifts the position of equilibrium to the left (by Le Chatelier's principle) to oppose the decrease [1 mark]. 2) More electrons are released on the electrode, making E(Zn²⁺/Zn) more negative [1 mark]. 3) Since E_cell = E(cathode) − E(anode), the gap between the two electrode potentials increases, so E_cell increases [1 mark].
Common Exam Questions (3)
What is oxidation?
Oxidation is the loss of electrons from a species, which corresponds to an increase in its oxidation state (oxidation number).
What is a standard hydrogen electrode (SHE)?
The standard hydrogen electrode (SHE) is the reference half-cell used to measure standard electrode potentials under standard conditions (298 K, 100 kPa H2 gas, 1.00 mol dm⁻³ H⁺(aq) with a platinum electrode), assigned a potential of exactly 0.00 V.
How do you calculate the standard cell potential (E°cell)?
Calculate the standard cell potential using E°cell = E°(reduction / right-hand positive electrode) - E°(oxidation / left-hand negative electrode).
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