Electron transfer reactions

Chemistry · Reactivity 3: What are the mechanisms of chemical change?. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Oxidation is loss of electrons, or a rise in oxidation state. Reduction is gain of electrons, or a fall in oxidation state. They always happen together. The oxidizing agent takes the electrons, so it is itself reduced; the reducing agent is itself oxidized.

A more reactive metal loses electrons more easily, so it displaces a less reactive metal from a solution of its ions. A voltaic cell uses such a reaction to produce electricity. An electrolytic cell does the opposite: electrical energy drives a reaction that would not happen by itself.

In organic chemistry, oxidation usually means gaining oxygen or losing hydrogen, and reduction means the reverse.

Rules to remember

Oxidation Is Loss of electrons, Reduction Is Gain (OIL RIG)
Half-equation in acid: balance the main atoms, add H₂O for O, add H⁺ for H, then add e⁻ to balance the charge
In every cell, oxidation is at the anode and reduction is at the cathode
Voltaic cell: the anode is negative, electrons flow through the wire from anode to cathode, chemical energy → electrical energy
Electrolytic cell: the anode is positive. Molten salt: metal forms at the cathode, non-metal at the anode
Oxidation with acidified K₂Cr₂O₇: primary alcohol → aldehyde (distil) → carboxylic acid (reflux); secondary alcohol → ketone; tertiary alcohol: no reaction
Reduction reverses this: aldehyde or carboxylic acid → primary alcohol, ketone → secondary alcohol. Alkene + H₂ → alkane

Common mistake

Students call the species that is oxidized the oxidizing agent. It is the other way round: the oxidizing agent is the one that gets reduced. They also swap the signs of the electrodes between voltaic and electrolytic cells.

Worked example

Deduce the half-equation for the reduction of Cr₂O₇²⁻ to Cr³⁺ in acidic solution.

Balance Cr: Cr₂O₇²⁻ → 2Cr³⁺.
Balance O with water, then H with H⁺: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O.
Charges: left is −2 + 14 = +12, right is 2 × (+3) = +6, so add 6e⁻ to the left.
Check with oxidation states: each Cr goes from +6 to +3, so 2 Cr atoms gain 6 electrons.

Answer: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

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Reactivity 3.2 Electron Transfer Reactions

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