Electron transfer reactions

A number tracking how many electrons an atom has **gained or lost** relative to the free element — the charge it would have if all bonds were ionic.

Always **0** (e.g. Na, O_{2}, Cl_{2}, S_{8}).

**Equal to its charge** (e.g. Mg²⁺ is +2, Cl⁻ is −1).

Oxygen is **−2**; hydrogen is **+1** — except peroxides (O is −1) and metal hydrides (H is −1).

They add up to the **total charge**: 0 for a neutral compound, the ion charge for a polyatomic ion.

An **increase** in oxidation state — the atom has **lost** electrons (OIL).

A **decrease** in oxidation state — the atom has **gained** electrons (RIG).

**O**xidation **I**s **L**oss, **R**eduction **I**s **G**ain (of electrons).

The species that **takes** electrons and is itself **reduced** (its oxidation state goes down), e.g. O_{2}, Cl_{2}.

The species that **gives** electrons and is itself **oxidised** (its oxidation state goes up), e.g. a reactive metal.

An atom's oxidation state **changes** during the reaction — so electrons have been transferred.

**+6** — four O at −2 (= −8) plus S must equal the ion charge −2, so S = +6.

An equation showing **just the oxidation or just the reduction** part of a redox reaction, with the electrons (e⁻) included.

On the **right** (product) side — oxidation is **loss** of electrons.

On the **left** (reactant) side — reduction is **gain** of electrons.

**O**xidation **I**s **L**oss, **R**eduction **I**s **G**ain (of electrons).

Balance the **atoms** first, then add **electrons** to the more positive side so the **charge** balances.

$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-}$ — an oxidation (loses 2e⁻).

$\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}$ — a reduction (gains 2e⁻).

**Multiply** so both transfer the same number of electrons, then **add** them and **cancel** the e⁻.

So the **electrons lost equal the electrons gained** — they must cancel exactly in the overall equation.

Both the **atoms** and the **total charge** must balance, with **no electrons** left over.

$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ — both halves transfer 2e⁻, which cancel.

A ranking of metals from **most reactive** (top) to **least reactive** (bottom), by how readily they lose electrons.

When a **more reactive** metal pushes a **less reactive** metal out of a solution of its ions.

A **more reactive** metal **displaces** a less reactive metal from a solution of its ions.

Electrons are **transferred**: the metal is **oxidised** (loses e⁻) and the metal ion is **reduced** (gains e⁻).

**O**xidation **I**s **L**oss, **R**eduction **I**s **G**ain — of electrons.

Oxidation: Zn → Zn²⁺ + 2e⁻; Reduction: Cu²⁺ + 2e⁻ → Cu.

Metals **above hydrogen** in the series → give a **salt + hydrogen gas**. Cu, Ag (below H) do not.

**Salt + hydrogen** (e.g. Mg + 2HCl → MgCl_{2} + H_{2}).

The most reactive ones (K, Na, Ca) → **metal hydroxide + hydrogen** (e.g. 2Na + 2H_{2}O → 2NaOH + H_{2}).

Add each metal to the **other's salt solution**; the metal that reacts (displaces) is the **more reactive**.

A **reducing agent** — it donates electrons (and is itself oxidised).

A **colour change** of the solution and a **deposit** of the displaced metal on the added metal.

A device that links a **redox reaction** to a flow of electrons through a wire — either making electricity (voltaic) or driven by it (electrolytic).

A cell in which a **spontaneous** redox reaction converts **chemical energy into electrical energy** (a battery).

A cell in which an external power supply drives a **non-spontaneous** reaction — **electrical energy into chemical energy** (electrolysis).

**Oxidation** (loss of electrons) — remember **AN OX**.

**Reduction** (gain of electrons) — remember **RED CAT**.

Anode = **negative (−)**, cathode = **positive (+)**.

Anode = **positive (+)**, cathode = **negative (−)** — the opposite of a voltaic cell.

Always from the **anode to the cathode** (in both cell types).

Completes the circuit and keeps each half-cell **neutral**: anions move toward the anode, cations toward the cathode.

Voltaic = **spontaneous**, makes electricity; electrolytic = **driven** by a supply, uses electricity.

$\text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag}(s)$ — reduction (gain of one electron).

The reference half-cell, assigned **E° = 0.00 V**, against which all other standard electrode potentials are measured.

The voltage of a half-cell measured against the **SHE** under **standard conditions** (298 K, 1 mol dm⁻³, 100 kPa).

**298 K**, solution concentrations **1 mol dm⁻³**, and gas pressures **100 kPa**.

As **reductions** — electrons on the left (e.g. Cu²⁺ + 2e⁻ ⇌ Cu).

The species is reduced more readily — it is a **stronger oxidising agent**.

The species is oxidised more readily — it is a **stronger reducing agent**.

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$ — given in the data booklet.

The one with the **more positive E°** — it is reduced.

The cell reaction is **spontaneous (feasible)** as written (and ΔG° is negative).

The reaction is **not spontaneous** in that direction (ΔG° is positive).

Forgetting that −(negative) **adds**: E°cell = E°(cathode) − E°(anode), so put brackets around each value.

$\Delta G^{\circ} = -nFE^{\circ}_{cell}$ — a positive E°cell makes ΔG° negative, the condition for spontaneity (F = 96 500 C mol⁻¹).

The species **reduced at the cathode** (it gains electrons); the species oxidised at the anode is the reducing agent.

Using an **external power supply** to drive a **non-spontaneous** redox reaction in a molten or aqueous electrolyte.

It is **negative**; **cations are reduced** there (gain electrons).

It is **positive**; **anions are oxidised** there (lose electrons). Remember **PANIC** — Positive Anode Is oxidation.

The **metal** at the cathode and the **non-metal** at the anode (e.g. molten NaCl → Na + Cl₂).

In an **aqueous** electrolyte, **water competes** with the ions, so only **one** species is discharged at each electrode.

**Position in the E° series**, **ion concentration**, and the **electrode material / overpotential**.

**Hydrogen gas** — water is reduced (2H₂O + 2e⁻ → H₂ + 2OH⁻) because the metal ion is too reactive.

**Chlorine gas** — a concentrated halide is discharged in preference to O₂.

$Q = I\,t$ — charge (C) = current (A) × time (s). NOT in the data booklet.

$n(e^{-}) = \dfrac{Q}{F}$ with $F = 96\,500$ C mol⁻¹ (given in the booklet).

The charge carried by **one mole of electrons**, **F = 96 500 C mol⁻¹**.

**Q = It → n(e⁻) = Q/F → ÷ electrons in the half-equation → moles of product → m = nM**.