Redox reactions as electron transfer

What is redox in terms of electrons?

In plain words: when sodium burns in chlorine, each sodium atom hands one electron to chlorine, giving Na⁺ and Cl⁻ ions. Sodium, which loses the electron, is oxidised; chlorine, which gains it, is reduced. Because an electron cannot vanish, one substance can lose electrons only if another takes them, which is why oxidation and reduction always go together.

From the classical idea to electron transfer

By the classical idea, sodium is oxidised in each of these reactions, because oxygen or a more electronegative element is added to it, while chlorine, oxygen and sulphur are reduced, because the electropositive element sodium is added to them (NCERT Section 7.2, page 237):

• 2Na(s) + Cl₂(g) → 2NaCl(s)
• 4Na(s) + O₂(g) → 2Na₂O(s)
• 2Na(s) + S(s) → Na₂S(s)

The Na₂O equation is the specified oxide-formation example, not a prediction that burning sodium always gives only this oxide; peroxide can form too. All three specified products are ionic compounds (see the ionic bond), better written Na⁺Cl⁻, (Na⁺)₂O²⁻ and (Na⁺)₂S²⁻. The charges show what really happened: each sodium atom loses one electron and the non-metal takes them. Two electrons move in 2Na + Cl₂ and in 2Na + S, but four move in 4Na + O₂, because each of the two oxygen atoms takes two.

Half reactions

For the formation of sodium chloride, the two steps are (page 237):

Each step is a half reaction: it shows the electrons explicitly. On adding the two, the 2e⁻ on each side cancel, because every electron lost is gained. The displayed state-free half reactions are electron bookkeeping. If gaseous ions are explicitly used as NCERT does, lattice formation is an additional step needed to reach the solid; states cannot be ignored in a thermochemical or electrode-potential calculation.

Sodium, which is oxidised, is the reducing agent: it donates electrons and so reduces chlorine. Chlorine, which is reduced, is the oxidising agent: it accepts electrons and so oxidises sodium. NCERT points out that the electron definitions were reached by matching each species’ classical behaviour with its electron transfer (page 237), so the two views agree wherever both can be used.

The electron view also settles the case that troubled the classical idea, 2Na + H₂ → 2NaH: sodium hydride is ionic, Na⁺H⁻, so sodium loses electrons and hydrogen gains them (NCERT Problem 7.2, Example 1 below).

Competitive electron transfer reactions

Zinc in copper nitrate solution. Place a strip of zinc in aqueous copper nitrate for about an hour. The strip becomes coated with reddish copper and the blue colour of the solution, due to Cu²⁺, fades away (NCERT Section 7.2.1, Fig. 7.1, page 238). Passing hydrogen sulphide through the colourless solution and making it alkaline with ammonia gives white zinc sulphide, ZnS, which shows that Zn²⁺ ions have formed. The reaction is

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

The nitrate ions take no part, so they are left out. Zinc loses two electrons and is oxidised; the copper ion gains them and is reduced.

The reverse is negligible under ordinary conditions. A strip of copper in zinc sulphate solution shows no visible change. Even the very sensitive test with hydrogen sulphide, which would give black copper sulphide, CuS, finds no Cu²⁺. So the equilibrium for the zinc–copper reaction lies very far towards Zn²⁺ and Cu (page 238).

Copper in silver nitrate solution. The solution slowly turns blue and silver deposits on the copper rod (NCERT Fig. 7.2, pages 238–239):

Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)

Copper releases two electrons, one to each of two silver ions, and the equilibrium again greatly favours the products.

Cobalt in nickel sulphate solution. For Co(s) + Ni²⁺(aq) ⇌ Co²⁺(aq) + Ni(s), tests at equilibrium find both Ni²⁺ and Co²⁺ at moderate concentrations: neither side is greatly favoured, because cobalt and nickel release electrons about equally readily (pages 238–239).

Electron-releasing tendency and the activity series

Zinc releases electrons to copper ions, and copper releases electrons to silver ions, so the tendency of these metals to release electrons is in the order Zn > Cu > Ag (page 239). Under comparable aqueous conditions, a higher metal is thermodynamically favoured to reduce ions of a lower one. Concentrations, complex formation, oxide films and competing reactions with water can change observed behaviour; the series predicts tendency, not rate. Careful: in NCERT’s table of standard electrode potentials (Table 7.1, page 251), the best electron releasers, such as Li, K and Na, are at the bottom, so there a metal reduces the ions of metals listed above it. NCERT compares this competition for electrons with the competition among acids for releasing protons. Extended to many metals, the list becomes the metal activity series, or electrochemical series. The same competition is used in galvanic cells, in which a redox reaction is the source of electrical energy (page 239).

Solved examples

Common mistakes to avoid

Quick revision summary

• Oxidation: loss of electrons; reduction: gain of electrons.
• Oxidising agent: electron acceptor (is reduced); reducing agent: electron donor (is oxidised).
• Half reactions show the electrons; electrons lost = electrons gained.
• Zn + Cu²⁺ and Cu + 2Ag⁺ go almost completely; Co + Ni²⁺ gives a mixture.
• Electron-releasing tendency: Zn > Cu > Ag; extended, this is the activity (electrochemical) series.

Questions students ask

Why is oxidation called loss of electrons?

Oxidation is called loss of electrons because combining with oxygen, the original meaning of oxidation, takes electrons away from the other element: when aluminium burns, 4Al + 3O₂ → 2Al₂O₃, each aluminium atom gives three electrons to the more electronegative oxygen and becomes Al³⁺. Chemists saw the same loss of electrons when metals react with chlorine or sulphur instead, so the name oxidation was extended to every loss of electrons, with or without oxygen, and reduction to every gain. The word is historical: the modern definition counts the electrons, not the oxygen.

Is every chemical reaction a redox reaction?

No, a reaction is a redox reaction only if electrons are transferred or shifted between atoms, which shows up as a change in the oxidation number of at least one element; acid–base neutralisations and most precipitation reactions are not redox. In HCl + NaOH → NaCl + H₂O a proton passes from the acid to the base, and in AgNO₃ + NaCl → AgCl + NaNO₃ the ions only change partners: every element keeps its oxidation number (H +1, O −2, Na +1, Cl −1, Ag +1, N +5). A free element on either side usually signals a redox reaction, but the sure test is a change of oxidation number: 3O₂ → 2O₃, oxygen to ozone, is not redox.

Why are metals good reducing agents?

Many reactive metals are good reducing agents because forming their ions is favourable in the reaction medium; noble metals are much weaker. A typical oxidation is Ca → Ca²⁺ + 2e⁻. The alkali metals, with one ns electron and the lowest ionisation enthalpies of their periods, are among the strongest; by E^⊖ in water lithium leads (−3.05 V) despite its higher ionisation enthalpy, because its small Li⁺ ion is so strongly hydrated. Non-metals such as oxygen and the halogens, one or two electrons short of an octet, are typically oxidising agents, though hydrogen and carbon are good reducing agents.

Are electrons transferred or shared in a redox reaction?

Electron transfer can occur between ions or at electrodes without making a new ionic compound, for example Fe³⁺ + e⁻ → Fe²⁺. Formation of KBr is conveniently described by ionic electron bookkeeping, while N₂ + 3H₂ → 2NH₃ forms polar covalent bonds. Oxidation numbers assign their bonding electrons formally to the more electronegative atom; they are not measurements of complete physical charge transfer. Such reactions still count as redox, because chemists treat the shift as complete and track it with oxidation numbers: nitrogen goes from 0 to −3 and is reduced, and hydrogen from 0 to +1 and is oxidised.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Write the two half reactions for 4Na(s) + O₂(g) → 2Na₂O(s) and show that the electrons balance.

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   Oxidation: 4Na → 4Na⁺ + 4e⁻. Reduction: O₂ + 4e⁻ → 2O²⁻. Four electrons are lost and four gained; adding gives 4Na + O₂ → 2(Na⁺)₂O²⁻.

2. Q2. In 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), how many electrons are transferred, and which species is the oxidising agent?

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   Each Al loses 3e⁻, so 2Al lose 6e⁻; each Cu²⁺ gains 2e⁻, so 3Cu²⁺ gain 6e⁻. Six electrons are transferred. Cu²⁺ accepts them, so it is the oxidising agent.

3. Q3. A silver spoon is left in copper nitrate solution. Does the solution change? Explain.

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   No visible displacement is expected under ordinary conditions. Silver releases electrons less readily than copper (Zn > Cu > Ag), so reduction of Cu²⁺ by silver is unfavourable under comparable conditions; the reverse reaction, Cu + 2Ag⁺, is the one that goes.