Redox and Electrochemistry

This lesson covers oxidation states, redox reactions, half-equations, standard electrode potentials, electrochemical cells, electrolysis, Faraday's laws, disproportionation, and the effect of non-standard conditions. Electrochemistry connects chemical energy to electrical energy and is central to understanding batteries, fuel cells, electrolysis, and corrosion.

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Oxidation and Reduction

Key Definition: Oxidation is the loss of electrons (or an increase in oxidation state). Reduction is the gain of electrons (or a decrease in oxidation state). The two always occur together — a redox reaction.

Concept	Oxidation	Reduction
Electrons	Loss	Gain
Oxidation state	Increases	Decreases
Oxygen	Gain	Loss
Hydrogen	Loss	Gain

The mnemonic OILRIG helps: Oxidation Is Loss, Reduction Is Gain (of electrons).

An oxidising agent (oxidant) is a substance that causes oxidation by accepting electrons — it is itself reduced. A reducing agent (reductant) causes reduction by donating electrons — it is itself oxidised.

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Oxidation States (Oxidation Numbers)

Oxidation states are assigned using a set of rules:

1. Elements in their standard state have an oxidation state of 0.
2. Simple monatomic ions have an oxidation state equal to the charge on the ion (e.g. Na⁺ = +1, Cl⁻ = −1).
3. In compounds, fluorine is always −1.
4. Oxygen is usually −2 (except in peroxides where it is −1, and in OF₂ where it is +2).
5. Hydrogen is usually +1 (except in metal hydrides where it is −1).
6. The sum of oxidation states in a neutral compound is 0; in an ion, it equals the charge on the ion.

Using Oxidation States to Identify Redox

If the oxidation state of an element increases, it has been oxidised. If it decreases, it has been reduced.

Disproportionation

Key Definition: Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced — the element in a single oxidation state is converted into two different oxidation states.

Worked Example: Disproportionation of Chlorine in Sodium Hydroxide

When chlorine gas is bubbled into cold dilute sodium hydroxide:

Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Chlorine in Cl₂ has an oxidation state of 0.

In NaCl, chlorine has an oxidation state of −1 (reduced — gained one electron).

In NaClO (sodium chlorate(I)), chlorine has an oxidation state of +1 (oxidised — lost one electron).

Therefore chlorine has been both oxidised (0 → +1) and reduced (0 → −1) — this is a disproportionation reaction.

This reaction is the basis of household bleach. In hot concentrated NaOH, the product is NaClO₃ (sodium chlorate(V)) instead, where Cl has an oxidation state of +5.

Worked Example: Disproportionation of Copper(I) Oxide

Cu₂O can undergo disproportionation in acidic solution:

Cu₂O(s) + H₂SO₄(aq) → Cu(s) + CuSO₄(aq) + H₂O(l)

In Cu₂O, copper has oxidation state +1. In Cu(s), copper has oxidation state 0 (reduced). In CuSO₄, copper has oxidation state +2 (oxidised). This is disproportionation: Cu⁺ → Cu (reduced) and Cu⁺ → Cu²⁺ (oxidised).

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Half-Equations

Complex redox reactions can be broken down into two half-equations — one for oxidation and one for reduction. To balance a half-equation:

1. Balance the atoms other than O and H.
2. Balance O atoms by adding H₂O.
3. Balance H atoms by adding H⁺.
4. Balance charge by adding electrons.

For example, the reduction of dichromate(VI) ions in acidic solution:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

To combine two half-equations into an overall equation, multiply each half-equation so that the number of electrons is the same in both, then add and cancel the electrons.

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Standard Electrode Potentials (E⦵)

Key Definition: The standard electrode potential (E⦵) is the EMF of a half-cell measured under standard conditions (298 K, 100 kPa, 1.00 mol dm⁻³ solutions) relative to the standard hydrogen electrode, which is assigned a value of 0.00 V.

The SHE consists of:

• Hydrogen gas at 100 kPa bubbled over a platinised platinum electrode.
• A 1.00 mol dm⁻³ solution of H⁺ ions.
• Temperature of 298 K.

The platinum electrode serves two functions: it provides an inert conducting surface for the electrode reaction and the platinised (roughened) surface increases the surface area for adsorption of hydrogen.

The Electrochemical Series

Half-reactions are listed by convention as reductions: Mⁿ⁺ + ne⁻ → M

• Species with more positive E⦵ values are stronger oxidising agents (more easily reduced).
• Species with more negative E⦵ values are stronger reducing agents (more easily oxidised).

Half-reaction	E⦵ / V
F₂ + 2e⁻ → 2F⁻	+2.87
Cl₂ + 2e⁻ → 2Cl⁻	+1.36
Ag⁺ + e⁻ → Ag	+0.80
Cu²⁺ + 2e⁻ → Cu	+0.34
2H⁺ + 2e⁻ → H₂	0.00
Fe²⁺ + 2e⁻ → Fe	−0.44
Zn²⁺ + 2e⁻ → Zn	−0.76
Mg²⁺ + 2e⁻ → Mg	−2.37

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Electrochemical Cells

An electrochemical cell consists of two half-cells connected by a salt bridge and an external wire. The salt bridge allows ions to flow to maintain electrical neutrality and complete the circuit. It is typically made from a strip of filter paper soaked in KNO₃(aq) or a U-tube filled with KNO₃ solution (chosen because K⁺ and NO₃⁻ are spectator ions that do not participate in the cell reaction).

The EMF (electromotive force) of the cell is calculated as:

E⦵(cell) = E⦵(most positive half-cell) − E⦵(least positive half-cell)

or equivalently:

E⦵(cell) = E⦵(cathode) − E⦵(anode)

In a cell:

• Oxidation occurs at the negative electrode (anode).
• Reduction occurs at the positive electrode (cathode).
• Electrons flow from the anode to the cathode through the external circuit.

graph LR
    subgraph Anode["Anode (−)"]
        A1["Oxidation occurs<br/>Metal dissolves:<br/>M → M²⁺ + 2e⁻"]
    end
    subgraph Cathode["Cathode (+)"]
        C1["Reduction occurs<br/>Ions deposited:<br/>M²⁺ + 2e⁻ → M"]
    end
    A1 -->|"Electrons flow<br/>through external wire"| C1
    A1 <-->|"Salt bridge<br/>(ions flow to maintain<br/>electrical neutrality)"| C1

Worked Example: Calculating E⦵(cell)

Calculate the EMF of a cell made from a zinc half-cell and a copper half-cell under standard conditions.

Given: E⦵(Zn²⁺/Zn) = −0.76 V, E⦵(Cu²⁺/Cu) = +0.34 V

Solution:

Copper has the more positive E⦵, so it is the cathode (reduction occurs here): Cu²⁺ + 2e⁻ → Cu

Zinc has the more negative E⦵, so it is the anode (oxidation occurs here): Zn → Zn²⁺ + 2e⁻

E⦵(cell) = E⦵(cathode) − E⦵(anode)

E⦵(cell) = (+0.34) − (−0.76)

E⦵(cell) = +1.10 V

The overall cell reaction is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Cell notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Predicting Feasibility of Reactions

A reaction is thermodynamically feasible if E⦵(cell) is positive. However, a positive E⦵ does not guarantee the reaction will occur at a measurable rate — kinetic factors (such as high activation energy) may prevent it.