Electrochemistry

Oxidation and reduction always occur together — you cannot have one without the other. This paired process is called a redox reaction.

• Oxidizing agent — accepts electrons from another species; it gets reduced in the process.
• Reducing agent — donates electrons to another species; it gets oxidized in the process.

Term	Electrons	Oxidation state	Role in the other species
Oxidation	Loss of e⁻	Increases (goes more positive)	Species is the reducing agent
Reduction	Gain of e⁻	Decreases (goes more negative)	Species is the oxidizing agent

Oxidation states are a bookkeeping tool used to track which atoms gain or lose electrons. They are assigned numbers, not actual charges, but they behave like charges for bookkeeping purposes.

Rules for Assigning Oxidation States

Worked Example — Mn in KMnO₄

Worked Example — Cr in Cr₂O₇²⁻

Assign oxidation states to all atoms before and after the reaction. Then compare:

• Increase in OS → that atom (and its species) is oxidized
• Decrease in OS → that atom (and its species) is reduced
• No change in any OS → not a redox reaction

Example — Zn + CuSO₄ → ZnSO₄ + Cu

Atom	Before	After	Change	Conclusion
Zn	0	+2	+2	Oxidized (reducing agent)
Cu	+2	0	−2	Reduced (oxidizing agent)
S	+6	+6	0	Spectator
O	−2	−2	0	Spectator

The half-reaction method is the systematic approach. Electrons lost must equal electrons gained.

Steps (Acidic Solution)

1. Split the overall equation into an oxidation half-reaction and a reduction half-reaction.
2. Balance all atoms other than O and H.
3. Add H₂O molecules to balance oxygen atoms.
4. Add H⁺ ions to balance hydrogen atoms.
5. Add electrons (e⁻) to balance the charge on each side.
6. Multiply each half-reaction by a factor so the number of electrons is equal in both.
7. Add the two half-reactions together, cancelling the electrons.
8. Check: all atoms balanced, all charges balanced.

Worked Example — MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic)

Multiply Fe half ×5 to equalize electrons:

A galvanic cell converts the energy of a spontaneous redox reaction into electrical energy. The two half-reactions are physically separated into compartments called half-cells.

Cell Notation

Example for Zn/Cu cell:  Zn | Zn²⁺(aq) || Cu²⁺(aq) | Cu

The double line (||) represents the salt bridge.

Every half-reaction has a measurable tendency to proceed as a reduction. This is its standard reduction potential (E°), measured at 25°C, 1 M concentrations, 1 atm, relative to the standard hydrogen electrode (SHE = 0.00 V).

Example — Zn/Cu Cell

Activity Series (Selected, Most to Least Reactive)

Metal	Half-reaction	E° (V)	Reactivity
K	K⁺ + e⁻ → K	−2.93	Most reactive reducing agent
Na	Na⁺ + e⁻ → Na	−2.71
Mg	Mg²⁺ + 2e⁻ → Mg	−2.37
Al	Al³⁺ + 3e⁻ → Al	−1.66
Zn	Zn²⁺ + 2e⁻ → Zn	−0.76
Fe	Fe²⁺ + 2e⁻ → Fe	−0.44
H	2H⁺ + 2e⁻ → H₂	0.00	Reference (SHE)
Cu	Cu²⁺ + 2e⁻ → Cu	+0.34
Ag	Ag⁺ + e⁻ → Ag	+0.80
Au	Au³⁺ + 3e⁻ → Au	+1.50	Least reactive (best oxidizing agent)

Electrolysis uses external electrical energy to drive a non-spontaneous redox reaction. An external power source forces electrons in the non-spontaneous direction.

• Cathode (−): reduction still occurs (connected to negative terminal of battery)
• Anode (+): oxidation still occurs (connected to positive terminal of battery)
• Unlike galvanic cells, the anode is now the positive electrode

Applications

Application	What happens
Electroplating	Metal ions from solution deposit on object (cathode); anode dissolves to replenish solution
Aluminium production	Al³⁺ reduced at cathode from molten Al₂O₃
Copper refining	Impure Cu anode dissolves; pure Cu deposits at cathode
Water electrolysis	2H₂O → 2H₂ + O₂; H₂ at cathode, O₂ at anode; volume ratio 2:1

Faraday's Law (Quantitative Electrolysis)

Mistake	What to do instead
Confusing anode/cathode polarity	Anode = oxidation ALWAYS. In galvanic: anode is −. In electrolytic: anode is +.
Reversing OIL RIG	Oxidation Is Loss of electrons. Reduction Is Gain. Use the mnemonic every time.
E°cell = sum of potentials	E°cell = E°cathode − E°anode (subtract, don't add). The anode value is subtracted.
OS of O in peroxides = −2	In H₂O₂ and peroxide ions, O has OS = −1, not −2.
Forgetting to balance charge with e⁻	After balancing atoms in each half-reaction, add e⁻ to make charges equal on both sides.
Reducing agent "gets reduced"	The reducing agent is oxidized (it donates electrons). The oxidizing agent is reduced.
Salt bridge not needed	Without a salt bridge, the half-cell solutions become charged and current stops.

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