Lithium Cells: Reactions and Function Revision Notes | A Level AQA Chemistry

1. Why Lithium Cells Matter

Your phone, laptop, and electric car all run on the same fundamental chemistry: the lithium cell. Understanding how it works is not just useful for the exam — it connects electrode potential theory directly to real technology.

At A Level, you need to go beyond "lithium is in the battery" and explain the precise electrode reactions, the direction of electron flow, and why lithium is such a powerful choice of electrode material. This lesson builds that understanding from the ground up.

2. The Electrochemical Foundations

Before writing the electrode equations, it helps to recall what every electrochemical cell has in common. A cell converts chemical energy into electrical energy through a spontaneous redox reaction — one species is oxidised at the negative electrode, and another is reduced at the positive electrode.

The key memory hook here is AN OX, RED CAT: oxidation at the anode (negative electrode during discharge), reduction at the cathode (positive electrode during discharge). Electrons flow through the external circuit from negative to positive — from the lithium electrode towards the cobalt oxide electrode.

Lithium has a very negative standard electrode potential ($E^\circ = -3.04$ V vs the standard hydrogen electrode). This means lithium is an exceptionally strong reducing agent — it releases electrons readily. Paired with a positive electrode that has a much less negative (or positive) $E^\circ$, the resulting cell EMF is large, which is why lithium cells deliver high voltages compared to many alternatives.

3. The Electrode Reactions in a Lithium Cell

Now for the chemistry that AQA specifically requires you to know. A lithium cell has two electrodes: a lithium metal negative electrode and a cobalt(IV) oxide ($\text{CoO}_2$) positive electrode. During discharge, both electrodes undergo distinct half-reactions.

At the negative electrode, lithium metal is oxidised. Each lithium atom loses one electron and becomes a lithium ion. The half-equation is straightforward:

At the positive electrode, the lithium ions that have migrated through the electrolyte combine with the cobalt oxide lattice and the electrons arriving from the external circuit. The cobalt oxide acts as a host structure, incorporating the lithium ion:

Notice the notation $\text{Li}^+[\text{CoO}_2]^-$. The square brackets indicate that the lithium ion has been incorporated (intercalated) into the cobalt oxide lattice, and the overall species carries a negative charge. The cobalt oxide framework has been reduced — it has gained an electron — while the lithium ion remains as $\text{Li}^+$ within the structure.

4. The Overall Cell Reaction and Electron Flow

Combining the two half-equations gives the overall cell reaction. The electron produced at the negative electrode is consumed at the positive electrode, so it cancels out:

Worked example: Deriving the Overall Cell Reaction

Combine the two half-equations for the lithium cell to produce the overall cell equation.

1. $\text{Li}(s) \rightarrow \text{Li}^+ + \text{e}^-$ — Oxidation at the negative electrode — lithium loses one electron. [B1]
2. $\text{Li}^+ + \text{CoO}_2 + \text{e}^- \rightarrow \text{LiCoO}_2$ — Reduction at the positive electrode forms lithium cobalt oxide. [B1]
3. $\text{Li}(s) + \text{CoO}_2 \rightarrow \text{LiCoO}_2$ — The $\text{Li}^+$ and $\text{e}^-$ appear on both sides and cancel. The overall equation shows lithium metal reacting with cobalt oxide to form the intercalated product. [M1]
4. Left: 1 Li, 1 Co, 2 O, charge = 0. Right: 1 Li, 1 Co, 2 O, charge = 0. — Both sides balance in terms of atoms and overall charge. [A1]

Answer: $\text{Li}(s) + \text{CoO}_2 \rightarrow \text{LiCoO}_2$

During discharge, electrons travel through the external circuit from the negative electrode (lithium) to the positive electrode (cobalt oxide). Simultaneously, lithium ions migrate through the internal electrolyte in the same direction — from the negative electrode towards the positive electrode. This dual movement of electrons (external) and ions (internal) constitutes the complete circuit.

5. Why Lithium? Evaluating the Choice of Electrode Material

The choice of lithium as the negative electrode material is not arbitrary. Lithium has the most negative standard electrode potential of any metal commonly encountered at A-level ($E^\circ = -3.04$ V vs SHE). This means the cell EMF — calculated as $E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ — is maximised when lithium is the anode material, because the more negative the anode potential, the larger the difference between the two half-cell potentials.

Lithium also has an exceptionally low relative atomic mass ($A_r = 6.94$) and low density. This means a lithium cell stores a large amount of charge per unit mass — a property called specific energy density. For portable devices and electric vehicles, this is critical: you want maximum energy from minimum mass.

There is an important caveat to the feasibility argument. A positive $E^\circ_{\text{cell}}$ confirms thermodynamic feasibility — the reaction can release energy spontaneously. It does not guarantee the reaction proceeds at a useful rate. Kinetic factors, such as the conductivity of the electrolyte and the rate of ion intercalation into the cobalt oxide lattice, also determine practical cell performance.

6. Visualising the Cell: Structure and Ion Movement

Picture the lithium cell as two compartments separated by an electrolyte (typically a lithium salt dissolved in an organic solvent, or a solid polymer electrolyte). On the left, solid lithium metal forms the negative electrode. On the right, the cobalt oxide layer forms the positive electrode.

The intercalation process at the positive electrode is worth examining closely. The $\text{CoO}_2$ lattice has a layered structure with gaps between the layers. Lithium ions slot into these gaps during discharge, and the electrons simultaneously reduce the cobalt oxide framework. During charging (in rechargeable lithium-ion variants), this process reverses: lithium ions are extracted from the lattice and lithium metal is re-deposited at the negative electrode.

7. Exam Technique and Common Mistakes

AQA questions on lithium cells tend to target three areas: writing the correct half-equations with the right notation, explaining the direction of electron and ion flow, and evaluating why lithium is a suitable electrode material. Each of these has specific traps that cost marks.

AQA may provide standard electrode potential data and ask you to calculate the cell EMF or assess feasibility. The method is consistent with what you have already seen in the electrode potentials topic — the lithium cell is a specific application of that general framework.

8. Numerical Application: Cell EMF Calculation

Worked example: Calculating Cell EMF for a Lithium Cell

Given that the standard electrode potential for the $\text{Li}^+/\text{Li}$ half-cell is $-3.04$ V and the standard electrode potential for the $\text{CoO}_2/\text{Li}^+[\text{CoO}_2]^-$ half-cell is approximately $+0.56$ V, calculate the standard cell EMF and state whether the cell reaction is thermodynamically feasible.

1. Cathode (positive electrode, reduction): $\text{CoO}_2$ half-cell, $E^\circ = +0.56$ V Anode (negative electrode, oxidation): $\text{Li}^+/\text{Li}$ half-cell, $E^\circ = -3.04$ V — The more positive $E^\circ$ is the cathode (reduction). The more negative $E^\circ$ is the anode (oxidation). [B1]
2. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ — This is the standard AQA formula for cell EMF. [M1]
3. $E^\circ_{\text{cell}} = (+0.56) - (-3.04)$ — Subtracting a negative number is equivalent to adding its positive value. Take care with the sign here — this is a common arithmetic error. [M1]
4. $E^\circ_{\text{cell}} = 0.56 + 3.04 = 3.60$ V — Adding 0.56 and 3.04: $0.56 + 3.04 = 3.60$ V. [A1]
5. $E^\circ_{\text{cell}} = +3.60$ V (positive value) — A positive cell EMF confirms the reaction is thermodynamically feasible under standard conditions. The large value reflects lithium's very negative electrode potential. [B1]

Answer: $E^\circ_{\text{cell}} = +3.60$ V; the reaction is thermodynamically feasible under standard conditions.

This large cell EMF of approximately 3.6 V is one reason lithium cells are so widely used. For comparison, a standard alkaline cell produces around 1.5 V. The combination of high voltage and low electrode mass makes lithium cells the dominant technology for portable energy storage — a direct consequence of the electrode chemistry you have now studied in full.