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Current collectorsStandard potentialsAlF₃ passivationOver-discharge damageComposite foils

This is one of the cleanest illustrations of periodic reasoning in the whole cell, and it is routinely misexplained as a cost decision — copper is expensive, aluminium is cheap, so aluminium goes where it can.

That explanation is wrong, and expensively so. The assignment is electrochemical. Swap the foils and one of them dissolves while the other disintegrates.

8.1The rule, and why it is not negotiable

A current collector is the metal foil that carries electrons between the active material coating and the external terminal. It stores no energy and does no chemistry — it just has to survive.

Important

Aluminium on the cathode, copper on the anode. They cannot be swapped. Not for cost reasons, not for weight reasons, and not with better engineering.

8.2Why not copper on the cathode

E°(Cu²⁺/Cu) = +0.34 V vs SHE, which is +3.45 V vs Li/Li⁺.

A cathode operates at 3.5 to 4.3 V vs Li/Li⁺ — above copper’s dissolution potential. Copper foil at cathode potential would simply dissolve as Cu²⁺ into the electrolyte, migrate to the anode, and plate out there, causing internal shorts.

8.3Why aluminium survives at the cathode

Aluminium is thermodynamically even more reactive than copper, at E° = −1.66 V vs SHE. By thermodynamics alone it should corrode instantly. It does not, and the reason is kinetic rather than thermodynamic.

In LiPF₆-based electrolytes, aluminium forms a passivating AlF₃ and Al₂O₃ surface layer that is insoluble, dense and self-healing. That film blocks further corrosion up to about 4.5 V.

Why this matters

Note the dependence on the salt. In LiTFSI-based electrolytes without additives, aluminium corrodes badly above about 3.8 V, because the TFSI⁻ anion does not supply the fluoride needed to build AlF₃.

This is a real, practical constraint on electrolyte reformulation, and a good example of why “just switch the salt” is never simple — the salt is holding the current collector together.

8.4Why not aluminium on the anode

Aluminium alloys with lithium below about 0.3 V vs Li/Li⁺, forming LiAl and related phases. A graphite anode operates at 0.1 V — well inside the alloying region.

The foil would lithiate, expand dramatically for the reasons given for silicon in chapter 6, embrittle, and disintegrate.

8.5Why copper works at the anode

Copper’s noble character — its positive standard potential of +0.34 V — is exactly what makes it stable in the strongly reducing environment at the anode. It does not alloy with lithium at any accessible potential.

It also has the second-highest electrical conductivity of any metal at 5.96 × 10⁷ S/m, behind only silver.

The trade-off, summarised

AluminiumCopper
PositionGroup 13, period 3Group 11, period 4 (d-block)
Density (g/cm³)2.708.96
Conductivity (×10⁷ S/m)3.55.96
Conductivity per unit massHigherLower
CostLowerHigher
Stable at cathode potential?Yes — via AlF₃ passivationNo — dissolves
Stable at anode potential?No — alloys with lithiumYes

In plain English

Read the last two rows together. Aluminium is lighter, cheaper and better per kilogram, and it still cannot go on the anode. Copper is heavier, dearer and worse per kilogram, and it still cannot go on the cathode. Neither metal can do the other’s job at any price.

POTENTIAL vs Li/Li⁺0 V1.02.03.04.05.0COPPER — STABLEDISSOLVES AS Cu²⁺ (E° = +0.34 V vs SHE)ALUMINIUM — PASSIVATED BY AlF₃ALLOYSWITH Li< 0.3 VGRAPHITE ANODE 0.1 VCATHODE 3.5–4.3 VThe overlap is empty. There is no single metal that survives both ends, which is why every lithium-ion cell contains two different foils.
Figure 8.1And this is why over-discharge is permanently destructive. Take a cell below about 1.5 V and the anode potential rises past copper's dissolution potential — the current collector itself goes into solution. On recharge the copper redeposits as conductive bridges. The cell may pass a bench test and short weeks later, which is why a deeply over-discharged cell should be replaced rather than revived. Note also that aluminium's survival is kinetic, not thermodynamic: it depends on the AlF₃ film that LiPF₆ supplies. Switch to LiTFSI without additives and aluminium corrodes above 3.8 V.

8.6Why over-discharge is permanently destructive

Important

If a cell is taken below roughly 1.5 V, the anode potential rises above copper’s dissolution potential — and the current collector itself dissolves. Copper ions then redeposit on recharge as conductive dendrites.

The insidious part is the delay. The cell may pass a capacity test immediately afterwards and short weeks later, once the dendrites have grown far enough. This is why a deeply over-discharged cell should be retired rather than nursed back — the damage is in a component that no capacity measurement inspects.

Technical framing

It is worth noticing that this is the same mechanism as section 8.2, arrived at from the other direction. Copper dissolves whenever it finds itself above +3.45 V vs Li/Li⁺, and there are two ways to arrange that: put it at the cathode, or drop the anode potential far enough that the anode becomes the high-potential electrode.

8.7Thin foils and composite collectors

Copper foil is typically 6 to 10 µm thick and represents a significant fraction of cell mass and cost while storing no energy at all. Two directions are being pursued.

  • Thinner copper, down toward 4.5 µm, which cuts mass and cost directly at the price of handling difficulty and higher sheet resistance.
  • Composite collectors — a few micrometres of polymer film metallised on both sides. These cut more mass again, and add a genuine safety property: the thin metal layer acts as a fuse if an internal short develops, limiting the current the short can draw.

Quick check: test yourself

1.A supplier proposes switching from LiPF₆ to LiTFSI to avoid HF generation. What breaks?

Show answer
The aluminium current collector. TFSI⁻ does not supply the fluoride needed to build the passivating AlF₃ film, so aluminium corrodes badly above about 3.8 V — below normal cathode operating potential. The salt is not just an ion source; it is maintaining the collector.

2.Why is a cell that was over-discharged once and then recharged still dangerous?

Show answer
Because below ~1.5 V the anode potential rises above copper’s +3.45 V vs Li/Li⁺ dissolution potential and the copper foil dissolves. Those ions redeposit on recharge as conductive dendrites, which may take weeks to grow into a short — so the cell can pass a capacity test and fail later.

3.Aluminium is lighter, cheaper and better per kilogram than copper. Why is it not used on both sides?

Show answer
Because it alloys with lithium below about 0.3 V and the anode operates at 0.1 V, so the foil would lithiate, expand, embrittle and disintegrate. The assignment is set by standard electrode potential, not by material properties or cost.

Chapter summary

Frequently asked questions

Why is copper used on the anode and aluminium on the cathode?+

Because of their standard electrode potentials, not their prices. Copper sits at +0.34 V vs SHE, which is +3.45 V vs Li/Li⁺ — below the 3.5 to 4.3 V a cathode operates at — so copper foil at cathode potential would dissolve as Cu²⁺, migrate to the anode and plate out, causing internal shorts. Aluminium cannot go on the anode because it alloys with lithium below about 0.3 V vs Li/Li⁺, and graphite operates at 0.1 V, so the foil would lithiate, expand, embrittle and disintegrate. Each metal is stable in exactly one of the two environments.

Why does aluminium survive at 4.2 V when it is more reactive than copper?+

Kinetics, not thermodynamics. Aluminium’s E° is −1.66 V vs SHE, so by thermodynamics alone it should corrode instantly. In LiPF₆-based electrolytes it forms a passivating AlF₃ and Al₂O₃ surface layer that is insoluble, dense and self-healing, and that film blocks further corrosion up to about 4.5 V. Note the dependence on the salt: in LiTFSI-based electrolytes without additives, aluminium corrodes badly above about 3.8 V because the TFSI⁻ anion does not supply the fluoride needed to build AlF₃. This is a real constraint on electrolyte reformulation.

Why is over-discharging a lithium cell permanently destructive?+

Because it dissolves the copper current collector. If a cell is taken below roughly 1.5 V, the anode potential rises above copper’s dissolution potential and the foil itself begins to dissolve. Copper ions then redeposit on recharge as conductive dendrites. The insidious part is the delay: the cell may pass a capacity test immediately afterwards and short weeks later, which is why deeply over-discharged cells should be retired rather than recovered.

Why is the industry moving to thinner and composite current collectors?+

Because copper foil at 6 to 10 µm is a significant fraction of cell mass and cost while storing no energy. Thinner copper down to about 4.5 µm cuts that directly. Composite collectors — a few micrometres of polymer film metallised on both sides — cut more mass again and add a safety property: the thin metal layer acts as a fuse if an internal short develops, limiting the current the short can draw.

Reviewed by

SG

Sahil Goyal

Co-founder, Wingzman

LinkedIn
SG

Sourabh Goyal

Co-founder, Wingzman

LinkedIn

The Periodic Table of the EV is an original educational series on the materials science of electric vehicles. All values are standard-condition literature figures for representative materials, not measured data from a specific product, and sources differ on several of them. Always verify against the specific material datasheet in use before making design, purchasing or certification decisions.