Classical idea of oxidation and reduction
Long ago, chemists looked only at oxygen and hydrogen.
- Oxidation = adding oxygen, or removing hydrogen. Example: 2Mg + O₂ → 2MgO (Mg gains O).
- Reduction = removing oxygen, or adding hydrogen. Example: CuO + H₂ → Cu + H₂O (CuO loses O).
Later the idea grew. Adding any electronegative element (an element that pulls electrons strongly, like Cl or F) is also oxidation. Removing a metal-like (electropositive) element is also oxidation. Example: 2K₄[Fe(CN)₆] + H₂O₂ → 2K₃[Fe(CN)₆] + 2KOH (K is removed, so the complex is oxidised).
In H₂S + Cl₂ → 2HCl + S, sulphur loses hydrogen (oxidised) and chlorine gains hydrogen (reduced). One cannot happen without the other.
Try it: at step 1 and step 2 of the 3D, watch which atom gains O and which atom loses H.
Oxidising agent and reducing agent
The oxidising agent (oxidant) makes another substance oxidised. It gets reduced itself.
The reducing agent (reductant) makes another substance reduced. It gets oxidised itself.
In CuO + H₂ → Cu + H₂O, H₂ is the reducing agent and CuO is the oxidising agent.
Redox in terms of electron transfer
Look at 2Na + Cl₂ → 2NaCl. There is no oxygen and no hydrogen, yet it is a redox reaction.
- Na → Na⁺ + e⁻ : sodium loses an electron. This is oxidation.
- Cl₂ + 2e⁻ → 2Cl⁻ : chlorine gains electrons. This is reduction.
Each of these lines is a half reaction. Add the two halves (after making electrons equal) and you get the whole reaction. The electrons lost must equal the electrons gained.
So: oxidation = loss of electrons, reduction = gain of electrons. The electron giver is the reducing agent. The electron taker is the oxidising agent.
Try it: at step 3 of the 3D, follow the single yellow electron from Na to Cl.
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Dip a zinc rod in blue copper sulphate. After some time the rod gets a brown coat of copper and the blue colour fades.
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Zinc gives electrons to Cu²⁺ ions. The opposite (copper rod in zinc sulphate) does not happen. So zinc gives away electrons more easily than copper.
In the same way, copper gives electrons to Ag⁺, so a copper wire in silver nitrate gets silver crystals and the solution turns blue. Zinc can give electrons to H⁺ of an acid (H₂ gas forms) but copper cannot.
This "competition for electrons" gives an order: Zn > Cu > Ag in giving electrons. Later this becomes the electrochemical series.
Try it: at step 5 of the 3D pick "Cu rod + Zn²⁺". Predict first. Then pick "Cu rod + Ag⁺".
Key formulas and definitions
- Oxidation: loss of electrons, e.g. Zn → Zn²⁺ + 2e⁻
- Reduction: gain of electrons, e.g. Cu²⁺ + 2e⁻ → Cu
- Redox = oxidation half + reduction half (electrons lost = electrons gained)
- Oxidising agent: takes electrons, gets reduced
- Reducing agent: gives electrons, gets oxidised
Worked examples
1. In 2Mg + O₂ → 2MgO, which is oxidised and which is the oxidising agent?
Mg gains oxygen, so Mg is oxidised. O₂ causes this, so O₂ is the oxidising agent (it is reduced).
2. In CuO + H₂ → Cu + H₂O, name the substance oxidised and the substance reduced.
H₂ gains oxygen → oxidised. CuO loses oxygen → reduced. H₂ is the reducing agent.
3. Write the two half reactions for 2Na + Cl₂ → 2NaCl.
Oxidation: 2Na → 2Na⁺ + 2e⁻. Reduction: Cl₂ + 2e⁻ → 2Cl⁻. Electrons lost (2) = electrons gained (2).
4. Will a reaction happen if an iron nail is put in copper sulphate solution? Write it.
Yes. Iron is more active than copper. Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s). Fe is oxidised; Cu²⁺ is reduced.
5. Silver nitrate solution is stored in a copper vessel. What happens?
Copper gives electrons to Ag⁺: Cu + 2Ag⁺ → Cu²⁺ + 2Ag. The vessel slowly dissolves and the solution turns blue. So it is a bad idea.
Common mistakes
- Thinking the oxidising agent is oxidised. It is the opposite: the oxidising agent gets reduced.
- Believing redox needs oxygen. 2Na + Cl₂ → 2NaCl has no oxygen but is redox.
- Writing a half reaction with electrons on the wrong side. In oxidation, electrons are products.
- Thinking any metal can push out any other metal. Only a more active metal can.