Theory — Electron Transfer in Redox

Theory — Electron Transfer and Energy

Oxidation and reduction

A redox reaction is one in which electrons are transferred. Oxidation is the loss of electrons and reduction is the gain of electrons; a useful reminder is OIL RIG, oxidation is loss, reduction is gain. The species that loses electrons is the reducing agent, and the one that gains them is the oxidizing agent. We track the electrons with oxidation numbers, assigned by a short set of rules.

Rules for oxidation numbersA free element is 0. A monatomic ion equals its charge. Oxygen is usually −2, hydrogen usually +1. Group 1 metals are +1 and Group 2 are +2. The oxidation numbers in a neutral compound sum to 0; in an ion they sum to the ion charge.
An increase in oxidation number is oxidation; a decrease is reduction

Galvanic cells and cell potential

In a galvanic cell the two half-reactions are separated so the electrons travel through a wire, doing useful work. Oxidation happens at the anode and reduction at the cathode. Each half-reaction has a standard reduction potential; the half-cell with the higher value is reduced (the cathode), and the lower one is oxidized (the anode).

Standard cell potentialcell = E°cathode − E°anode
A positive E°cell means the cell reaction is spontaneous as written
Reduction half-reactionE° (V)
Ag⁺ + e⁻ → Ag+0.80
Cu²⁺ + 2e⁻ → Cu+0.34
2H⁺ + 2e⁻ → H₂0.00
Pb²⁺ + 2e⁻ → Pb−0.13
Ni²⁺ + 2e⁻ → Ni−0.25
Fe²⁺ + 2e⁻ → Fe−0.44
Zn²⁺ + 2e⁻ → Zn−0.76

Track the electrons

Oxidation numbers reveal which atom loses electrons (oxidized) and which gains them (reduced).

Anode and cathode

Oxidation happens at the anode; reduction happens at the cathode.

Cell potential

E°cell = E°cathode − E°anode; a positive value means the cell drives current on its own.

Apparatus

The equipment a real electrochemistry experiment uses to build a galvanic cell and measure its potential. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.

redox half-cells
Galvanic cell
Two half-cells joined by a salt bridge; electrons flow through the external wire.
Vmeasures cell voltage
Voltmeter
Measures the cell potential E°cell of the running cell.
0.00reads V, I, R
Digital multimeter
Confirms voltage and checks the direction of electron flow.
reagent solutions
Electrolyte solutions
Provide the metal-ion solutions for each half-cell.
holds solutions
Beaker
Holds each half-cell solution.
0.000 gmeasures mass
Analytical balance
Weighs the electrodes to track mass gained or lost.

Instructions — Running the Virtual Experiment

This is a predict, reveal, and compare lab. In every part you work out the answer yourself first, enter it, and only then does the simulation reveal the experimental value so you can check your work against it.

Part A — Oxidation Numbers (Oxidation Numbers tab)
1
Choose a compound and the element to assign. Using the rules, calculate the oxidation number of that element by hand, enter it (use a minus sign for negative values), and click Check. Work through at least five compounds.
Part B — Galvanic Cell and Cell Potential (Cell Potential tab)
1
Choose two electrode metals. Predict which one is oxidized (the anode), then calculate the cell potential from E°cathode − E°anode, enter both, and click Check. The voltmeter reading is then revealed for comparison.

Simulation — The Electrochemistry Bench

Virtual LabOxidation Numbers · Cell Potential
CompoundElementYour valueActual
Choose a compound, calculate, and check.

Assign the oxidation number

Target element: Mn
Oxidation number— hidden
Electrodes
Metal 1 (E°)Zn (−0.76 V)
Metal 2 (E°)Cu (+0.34 V)

Build the cell

Anode (oxidized)— hidden
Cathode (reduced)— hidden
Voltmeter reading E°cell— hidden

Team Questions

Question 1. In a redox reaction, what is the term for the loss of electrons? (one word)
Question 2. What is the oxidation number of manganese in KMnO₄? (use + and a number)
Question 3. At which electrode does oxidation occur in a galvanic cell? (one word)
Question 4. For a Zn and Cu cell, what is the standard cell potential? (in volts, two decimals)

Example Lab Report

A worked example showing the expected format and the calculate, reveal, and compare workflow.

Redox Reactions

Chemistry | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Objective

To assign oxidation numbers, identify the species oxidized and reduced in a galvanic cell, and find the standard cell potential from electrode reduction potentials.

Theory

Oxidation is loss of electrons and reduction is gain of electrons. In a galvanic cell oxidation occurs at the anode and reduction at the cathode, and the cell potential is E°cell = E°cathode − E°anode.

Oxidation-number balance · E°cell = E°cathode − E°anode

Results (worked example)

QuantityValue
Oxidation number of Mn in KMnO₄+7
Zn–Cu cell: anode (oxidized)Zn
Zn–Cu cell: cathode (reduced)Cu
E°cell (0.34 − (−0.76))1.10 V

Zinc has the lower reduction potential, so it is oxidized at the anode while copper is reduced at the cathode, giving a cell potential of 1.10 V.

Conclusion

Oxidation numbers identified the electron transfer, and the measured cell potential matched E°cathode − E°anode, confirming the predicted spontaneous direction of the cell.

Practice Questions

Show all work. Use the oxidation-number rules and E°cell = E°cathode − E°anode. Reduction potentials are in the Theory table.

Question 1
Find the oxidation number of chromium in K₂Cr₂O₇.
Hint: K is +1 (×2 = +2), O is −2 (×7 = −14); 2 + 2Cr − 14 = 0, so Cr = +6.
Question 2
For a cell made of silver and zinc, identify the anode and calculate E°cell.
Hint: Zn (−0.76) is oxidized at the anode, Ag (+0.80) is the cathode; E°cell = 0.80 − (−0.76) = 1.56 V.
Question 5 — Challenge
Explain why a positive cell potential always means a spontaneous cell reaction.
Hint: ΔG° = −nFE°; with n and F positive, a positive E° makes ΔG° negative, and a negative ΔG° means spontaneous.