How Redox Chemistry Shapes Alkaline and Lithium-Ion Batteries

An AA alkaline battery is labelled 1.5 V. A rechargeable lithium-ion battery is usually labelled about 3.7 V. Both produce electricity through redox reactions, yet one is designed to be used once, while the other can be charged and discharged hundreds of times.

Why? If both batteries rely on oxidation and reduction, why can't we simply force the chemical reactions in an alkaline battery to run backwards?

The answer lies in the materials used at each electrode, how the products of the reactions form, and whether those changes can be reversed without damaging the cell.

[IMAGE: A split everyday scene showing a single-use AA alkaline battery powering a TV remote alongside a rechargeable lithium-ion battery in a smartphone. Include subtle labels identifying their nominal voltages, 1.5 V and 3.7 V.]

This comparison reveals an important idea in electrochemistry: a battery's performance depends on much more than the redox reaction itself. Electrode structure, ion movement, voltage, reversibility, and the resources used to manufacture the cell all influence its design.

We'll compare two well-established technologies:

- **Primary cell:** an alkaline zinc-manganese dioxide battery, commonly used in AA and AAA batteries.
- **Secondary cell:** a lithium-ion battery using graphite and lithium cobalt oxide electrodes, commonly called an LCO battery.

The chemistry of these cells builds directly on the galvanic-cell and reduction-potential principles used throughout HSC electrochemistry.

## 1. How does a redox reaction produce electricity?

Imagine placing zinc metal into a solution containing copper(II) ions. Zinc atoms lose electrons, while copper(II) ions gain them.

The overall reaction is:

\[
\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}
\]

The reaction is spontaneous. Chemical energy is released as the reactants form more stable products.

However, if the zinc and copper ions react directly in the same container, much of that energy is released as heat rather than being transferred through an external circuit.

A galvanic cell solves this problem by separating the oxidation and reduction reactions.

[IMAGE: A scientifically accurate galvanic cell showing two separate half-cells, a zinc electrode in zinc ion solution and a copper electrode in copper ion solution, joined by an external wire containing a light bulb and a salt bridge. Show electrons travelling from zinc to copper through the wire and ions moving through the salt bridge. Label anode, cathode, and the direction of electron flow.]

The zinc electrode undergoes oxidation:

\[
\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}
\]

The copper electrode undergoes reduction:

\[
\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}
\]

Because the reactions occur in separate locations, electrons must travel through the external wire to reach the copper electrode. We can use that electron flow to power an electrical device.

Two processes must occur simultaneously:

- Electrons travel through the external circuit.
- Ions move through the electrolyte to maintain electrical neutrality.

The electrolyte provides an internal pathway for ionic movement. It does not provide a pathway for electrons.

A useful way to picture this is to imagine two separate routes through a transport network. Electrons travel along the external route, while ions move through the internal electrolyte route. Both are necessary to keep charge moving around the complete circuit.

The analogy has one limitation: ions do not carry the same electrons through the electrolyte. The two forms of charge movement occur through different parts of the cell.

### The anode and cathode

For a galvanic cell during discharge:

| Electrode | Reaction | Electron behaviour |
|---|---|---|
| Anode | Oxidation | Electrons are produced |
| Cathode | Reduction | Electrons are consumed |

In a discharging galvanic cell, the anode is negative and the cathode is positive.

Keep the reaction definitions fixed:

- Anode = oxidation.
- Cathode = reduction.

This rule applies to both battery technologies we are about to examine.

## 2. The primary cell: alkaline zinc-manganese dioxide

Think about an ordinary AA battery. Its label might say 1.5 V, but inside it is a carefully arranged collection of chemicals.

An alkaline battery contains three important components:

- Zinc powder, which acts as the negative electrode.
- Manganese dioxide, \(\ce{MnO2}\), which acts as the positive electrode.
- Potassium hydroxide, \(\ce{KOH}\), which acts as the alkaline electrolyte.

The zinc is mixed into a gel, while manganese dioxide is combined with conductive carbon. A separator keeps the electrodes apart while allowing ions to move through the electrolyte.

[IMAGE: Detailed cross-section of a cylindrical AA alkaline battery. Show the outer steel can acting as the positive current collector, the manganese dioxide and conductive carbon cathode mixture against the can, a porous separator, the central zinc powder gel anode, potassium hydroxide electrolyte, and the negative terminal connection. Label the positive and negative terminals and distinguish the electrode materials.]

### What happens during discharge?

When you connect the battery to a device, zinc begins losing electrons.

Why zinc?

Zinc is relatively easy to oxidise. Its atoms can release electrons, which then travel through the external circuit towards the manganese dioxide electrode.

**At the anode, zinc is oxidised:**

\[
\ce{Zn(s) + 2OH^-(aq) -> ZnO(s) + H2O(l) + 2e^-}
\]

Each zinc atom changes from oxidation state 0 to +2.

It releases two electrons.

Those electrons travel through the external circuit and eventually reach the manganese dioxide electrode.

**At the cathode, manganese dioxide is reduced:**

\[
\ce{2MnO2(s) + H2O(l) + 2e^- -> Mn2O3(s) + 2OH^-(aq)}
\]

Manganese changes from oxidation state +4 in \(\ce{MnO2}\) to an average oxidation state of +3 in \(\ce{Mn2O3}\).

It gains the two electrons released by zinc.

Notice that the half-equations involve hydroxide ions and water. When we add the equations, these species cancel.

**Overall cell reaction:**

\[
\ce{Zn(s) + 2MnO2(s) -> ZnO(s) + Mn2O3(s)}
\]

This is a simplified representation of the main discharge reaction. Actual alkaline battery chemistry involves additional solid phases and can become more complicated as discharge continues. The equations above provide a useful model for following the electron transfer.

The half-equations are consistent with the simplified alkaline-cell chemistry presented in the [University of Calgary Chemistry Textbook](https://chem-textbook.ucalgary.ca/version2/chapter-17-electrochemistry-introduction/batteries-and-fuel-cells/single-use-batteries/).

### Why does the electrolyte matter?

You might notice that potassium hydroxide does not appear in the overall reaction.

Does that mean the electrolyte is unnecessary?

Not at all.

The hydroxide ions are consumed at the zinc electrode and regenerated at the manganese dioxide electrode. They can move through the electrolyte from the cathode towards the anode, supporting the internal movement of charge.

The electrolyte also provides a medium with relatively high ionic conductivity.

Without it, the separated redox reactions could not sustain a useful electrical current.

### Why isn't an ordinary alkaline battery rechargeable?

Here is the important distinction.

The overall reaction can be written backwards on paper:

\[
\ce{ZnO(s) + Mn2O3(s) -> Zn(s) + 2MnO2(s)}
\]

But writing a reverse equation does not guarantee that a real battery can undergo that reaction successfully.

During discharge, the zinc electrode is converted into zinc-containing products, while manganese dioxide changes into other manganese oxide compounds.

To recharge the cell, an external power supply would need to reverse these changes. However, ordinary alkaline cells are not designed to restore the original electrode materials reliably.

Several problems can occur:

- Zinc may not redeposit evenly.
- Unwanted side reactions can produce gases.
- The electrode structure can deteriorate.
- Leakage of the alkaline electrolyte can damage the cell.

The result is that an ordinary primary alkaline battery cannot be safely and reliably restored to its original condition by applying an external voltage.

Some specialist alkaline cells are designed for limited rechargeability, but that does not make an ordinary single-use AA battery rechargeable. Never attempt to recharge a battery that is labelled non-rechargeable.

The key point is that a primary cell uses a reaction that is spontaneous during discharge, but its practical design does not support repeated reversal.

## 3. Voltage: why 1.5 V does not mean exactly 1.5 V

An alkaline AA battery is commonly labelled 1.5 V.

That number describes its nominal voltage. It does not mean the battery always supplies exactly 1.50 V.

To understand why, we need to distinguish standard cell potential from actual operating voltage.

### Standard cell potential

Standard cell potential, \(E^\circ_{\mathrm{cell}}\), describes the potential difference between two electrodes under standard conditions.

It is calculated using standard reduction potentials:

\[
E^\circ_{\mathrm{cell}}
=
E^\circ_{\mathrm{red,cathode}}
-
E^\circ_{\mathrm{red,anode}}
\]

Both values must be taken from a table of reduction potentials, even though oxidation occurs at the anode during discharge.

A positive standard cell potential indicates that the overall cell reaction is spontaneous under standard conditions.

For the simplified alkaline zinc-manganese dioxide reaction, a textbook reference value is approximately +1.43 V.

This is close to the familiar 1.5 V nominal rating, but the two values are not interchangeable.

### Actual operating voltage

Real batteries do not operate under perfectly standard conditions.

Their voltage changes with:

- The amount of reactant remaining.
- The composition of the electrolyte.
- Temperature.
- The rate at which current is drawn.
- The internal resistance of the cell.

When a battery supplies current, some of its potential difference is lost internally.

A simplified relationship is:

\[
V_{\mathrm{terminal}}=E_{\mathrm{cell}}-Ir
\]

where:

- \(V_{\mathrm{terminal}}\) is the voltage measured across the battery terminals, in volts (V).
- \(E_{\mathrm{cell}}\) is the cell's potential under its current chemical conditions, in volts.
- \(I\) is the current supplied, in amperes (A).
- \(r\) is the internal resistance, in ohms (\(\Omega\)).

This equation shows why a battery can have a higher voltage when disconnected than when powering a device.

### Worked example: voltage under load

An alkaline battery has an open-circuit potential of 1.43 V. Its internal resistance is 0.40 \(\Omega\). It supplies a current of 0.25 A to a device.

Calculate its terminal voltage.

**Step 1**: Identify the relationship between internal resistance and terminal voltage.

\[
V_{\mathrm{terminal}}=E_{\mathrm{cell}}-Ir
\]

**Step 2**: Substitute the given values.

\[
\begin{aligned}
V_{\mathrm{terminal}}
&=1.43-(0.25\times0.40)\\
&=1.43-0.10\\
&=1.33\ \mathrm{V}
\end{aligned}
\]

**Step 3**: Interpret the result.

The battery supplies approximately 1.33 V to the device under this load.

The difference between 1.43 V and 1.33 V is the voltage lost across the battery's internal resistance in this simplified model.

This is why the nominal voltage printed on a battery should never be treated as a guaranteed operating voltage.

04The rechargeable cell: lithium-ion batteries

Now consider a smartphone battery.

It contains a very different arrangement of materials. In the particular lithium-ion chemistry we’ll examine, the two electrodes are:

• Graphite, which stores lithium ions between its carbon layers.
• Lithium cobalt oxide, \(\ce{LiCoO2}\), which also stores lithium ions within its layered crystal structure.

The electrolyte is typically a lithium salt, such as \(\ce{LiPF6}\), dissolved in an organic solvent. A porous separator keeps the electrodes physically apart while allowing lithium ions to pass between them.

The secret is intercalation

Why can lithium-ion batteries be recharged when ordinary alkaline batteries cannot?

The answer involves a process called intercalation.

Intercalation occurs when ions enter the spaces within a solid material without completely destroying its underlying structure.

Think of graphite as a stack of thin carbon sheets. Lithium ions can enter the spaces between these sheets and remain stored there.

The layered structure of lithium cobalt oxide also provides locations where lithium ions can be inserted and removed.

This gives the battery a useful property: lithium ions can move between the two electrodes repeatedly without requiring the entire electrode materials to be rebuilt after every discharge.

The analogy of inserting objects between stacked sheets is useful, but the actual process involves changes in the crystal structure and electron distribution. It is not simply lithium ions sitting loosely between sheets of carbon.

What happens during discharge?

When the battery powers your phone, it undergoes spontaneous redox reactions.

The graphite electrode releases lithium ions and electrons.

At the graphite electrode, oxidation occurs:

\[ \ce{LiC6(s) -> 6C(s) + Li+(electrolyte) + e^-} \]

The lithium ion leaves the graphite structure and enters the electrolyte.

The electron travels through the external circuit, where it can provide electrical energy to the phone.

At the other electrode, lithium ions and electrons are accepted by the cobalt oxide material.

At the lithium cobalt oxide electrode, reduction occurs:

\[ \ce{CoO2(s) + Li+(electrolyte) + e^- -> LiCoO2(s)} \]

The two half-equations combine to give:

Overall discharge reaction:

\[ \ce{LiC6(s) + CoO2(s) -> 6C(s) + LiCoO2(s)} \]

These are simplified end-member equations. Real lithium-ion cells operate over a range of compositions, rather than converting completely between these two limiting structures.

The important feature is that the lithium ions move through the electrolyte, while the electrons travel through the external circuit.

They reach the same destination by different routes.

What is actually being oxidised and reduced?

Let’s examine the oxidation states.

In \(\ce{CoO2}\), oxygen has an oxidation state of −2.

Therefore:

\[ x+2(-2)=0 \]

\[ x=+4 \]

Cobalt has an oxidation state of +4.

In \(\ce{LiCoO2}\), lithium is +1 and oxygen is −2:

\[ (+1)+x+2(-2)=0 \]

\[ x=+3 \]

Cobalt has been reduced from +4 to +3.

The graphite electrode also undergoes oxidation as the lithiated graphite releases electrons.

An important detail is that lithium remains in the +1 oxidation state throughout this process. It moves between the electrodes, but it is not the species being oxidised or reduced.

Why can the reaction be reversed?

Now imagine your phone battery is almost flat.

You connect it to a charger. The charger supplies electrical energy, forcing the cell to undergo reactions in the opposite direction.

The lithium ions leave the cobalt oxide electrode and travel through the electrolyte towards the graphite electrode.

At the same time, the external power supply drives electrons towards the graphite electrode.

The charging reactions are therefore:

At the lithium cobalt oxide electrode:

\[ \ce{LiCoO2(s) -> CoO2(s) + Li+ + e^-} \]

At the graphite electrode:

\[ \ce{6C(s) + Li+ + e^- -> LiC6(s)} \]

Overall charging reaction:

\[ \ce{6C(s) + LiCoO2(s) -> LiC6(s) + CoO2(s)} \]

Notice that the charging reaction is the reverse of the discharge reaction.

There is an important energy distinction here.

During discharge, the redox reaction is spontaneous and produces electrical energy. During charging, an external power supply provides energy to drive the non-spontaneous reverse reaction.

The battery is not somehow generating energy while recharging itself. The charger is transferring energy into the battery’s chemical system.

This is why charging is an externally driven process, rather than another spontaneous galvanic reaction.

There is also a useful distinction in electrode terminology. During discharge, graphite is the anode and lithium cobalt oxide is the cathode. During charging, the reaction directions reverse, so the oxidation and reduction sites switch. If a question uses anode and cathode, always establish whether the battery is charging or discharging.

Why doesn’t repeated charging destroy the electrodes?

Intercalation makes repeated charging possible, but it doesn’t make the battery indestructible.

The electrode structures undergo changes as lithium ions enter and leave. Over many cycles, side reactions, structural changes, and deterioration of the electrolyte can reduce the amount of energy the battery stores.

This is why rechargeable batteries gradually lose capacity.

The battery is designed to make the reverse reaction possible, but the reversal is never perfectly efficient or completely free of material changes.

05Comparing battery performance

The chemical differences between these cells produce very different practical properties.

Higher voltage does not automatically mean longer battery life

A common mistake is to assume that a 3.7 V battery must last longer than a 1.5 V battery.

Voltage alone cannot tell us how long a battery will operate.

We also need to know its capacity, which measures how much electric charge it can deliver.

Electrical energy depends on both charge and voltage:

\[ W=QV \]

where \(W\) is electrical energy in joules, \(Q\) is charge in coulombs, and \(V\) is voltage in volts.

A battery with a higher voltage may store less total energy than a larger battery with a lower voltage.

This is why engineers consider energy density: the amount of energy stored per unit mass or volume.

Lithium-ion batteries have high energy density, making them particularly useful when a device needs substantial energy without becoming too heavy or bulky. The Australian Government identifies this property as a major reason lithium-ion technology is suitable for phones, electric bikes, and electric vehicles in its National Battery Strategy.

Rechargeability and cycle life

A primary alkaline battery is generally discarded once its useful reactants have been consumed.

A lithium-ion battery can undergo many discharge-charge cycles because its electrode materials support reversible lithium-ion movement.

However, cycle life depends on the particular chemistry and how the battery is operated.

For example, the Australian Government’s YourHome resource lists approximately 1,000-10,000 cycles across lithium-ion stationary battery systems. This is a broad range for home energy storage, not a guaranteed lifespan for every smartphone or electric vehicle battery.

Temperature, depth of discharge, charging rate, and battery design all influence how long a battery lasts.

See the YourHome battery performance information for further comparisons.

The practical advantage is clear: a rechargeable cell can provide electrical energy repeatedly without requiring the entire battery to be replaced after every discharge.

But that advantage comes with additional design requirements, including charging controls, protection against overcharging, and careful management of operating temperature.

06The environmental trade-offs

It would be tempting to conclude that lithium-ion batteries are automatically better for the environment because they can be recharged.

That conclusion is too simple.

Battery chemistry determines not only how electricity is produced, but also which resources are needed to manufacture the cell and what happens when it reaches the end of its life.

Raw materials and resource demand

Consider the materials in our two batteries.

Alkaline batteries use:

• Zinc.
• Manganese dioxide.
• Potassium hydroxide.
• Steel and other structural materials.

These are not impact-free materials. Mining, processing, manufacturing, and repeated replacement all have environmental costs.

However, this chemistry does not require lithium or cobalt.

The lithium cobalt oxide battery uses:

• Lithium.
• Cobalt.
• Graphite.
• Copper and aluminium current collectors.
• An organic liquid electrolyte.

Lithium-ion batteries offer excellent energy density, but their manufacture depends on mineral extraction and processing.

Cobalt is particularly relevant to our LCO example.

According to the US Geological Survey’s 2025 Mineral Commodity Summaries, the Democratic Republic of the Congo accounted for an estimated 76% of global mined cobalt production in 2024.

That concentration creates supply-chain risks and makes responsible sourcing an important part of battery manufacturing.

It is also important not to generalise this issue to every lithium-ion battery. Different lithium-ion chemistries use different cathode materials. For example, lithium iron phosphate batteries (LFP) do not use cobalt in their cathodes.

Lithium demand is also closely connected to battery manufacturing. The USGS estimated that batteries accounted for 87% of global lithium end use in 2024.

These figures show why battery technology is also a resource-management challenge.

The Australian Government’s National Battery Strategy considers the entire value chain, from mineral extraction and refining through manufacturing, reuse, and recycling.

Why recycling matters

Recycling can reduce the need for newly extracted materials.

It can also recover useful metals from batteries that would otherwise become waste.

However, recycling is not automatic. Batteries need to be collected, sorted, dismantled, and processed using suitable methods. Different battery chemistries require different recovery approaches.

Lithium-ion batteries also create a particular disposal concern.

If damaged or crushed, they can short-circuit and overheat, potentially causing fires in rubbish trucks and waste-processing facilities.

In NSW, the government advises that lithium-ion batteries must not be placed in household waste or kerbside recycling bins. Small, undamaged handheld batteries should be taken to an appropriate battery recycling collection point, with their terminals covered using clear adhesive tape to reduce the risk of accidental short circuits.

Damaged batteries require special handling and should not be placed in ordinary household bins.

See the NSW Government’s lithium-ion battery safety and recycling guidance.

Standard household batteries, including many AA and AAA alkaline cells, can also be taken to participating B-cycle collection points. Check the Service NSW battery recycling information for accepted battery types and disposal locations.

Which battery is more sustainable?

There isn’t one answer for every application.

A rechargeable lithium-ion battery can avoid repeated replacements, particularly in devices that use substantial amounts of energy. Its high energy density also means less battery mass is needed for a given amount of stored energy.

However, manufacturing lithium-ion cells requires materials such as lithium and, in the LCO chemistry examined here, cobalt. The battery also needs appropriate protection and an effective recycling pathway.

An alkaline battery uses a different set of resources and is relatively inexpensive, but it is designed for single use. Repeatedly replacing cells can create a substantial material and waste burden.

The better environmental choice depends on the complete service provided by the battery:

• How much energy does it deliver?
• How many times can it be used?
• How long does it last?
• What resources are required to manufacture it?
• Can its materials be recovered at the end of its life?

The important lesson is that rechargeable does not automatically mean environmentally harmless, just as single-use does not automatically mean unsuitable.

Good battery design balances chemical performance, practical use, resource demand, safety, and end-of-life recovery.

The resource figures above come from the USGS 2025 Mineral Commodity Summaries. For another explanation of the graphite and lithium cobalt oxide electrode reactions, see OpenStax: Batteries and Fuel Cells.

04Questions and solutions

Question 1

An alkaline zinc-manganese dioxide battery is discharging. During a particular stage, 0.00600 mol of zinc reacts.

(a) Write the balanced oxidation and reduction half-equations.

(b) Write the overall balanced cell reaction.

(c) Identify which element is oxidised and which is reduced.

(d) Calculate the amount, in moles, of electrons transferred.

Solution 1

Zinc is oxidised, manganese is reduced, and 0.0120 mol of electrons are transferred.

(a) Write the half-equations.

Oxidation at the anode:

\[ \ce{Zn + 2OH^- -> ZnO + H2O + 2e^-} \]

Reduction at the cathode:

\[ \ce{2MnO2 + H2O + 2e^- -> Mn2O3 + 2OH^-} \]

The electrons are already balanced, with two electrons appearing in each half-equation.

(b) Add the half-equations.

Cancel the electrons, hydroxide ions, and water.

\[ \ce{Zn + 2MnO2 -> ZnO + Mn2O3} \]

(c) Determine the oxidation state changes.

For zinc:

\[ \ce{Zn^0 -> Zn^2+} \]

Zinc increases its oxidation state from 0 to +2. It loses electrons and is oxidised.

For manganese:

\[ \ce{Mn^4+ -> Mn^3+} \]

Manganese decreases its oxidation state from +4 to +3. It gains electrons and is reduced.

(d) Calculate the electron transfer.

Each mole of zinc releases two moles of electrons.

\[ n(e^-)=2n(\ce{Zn}) \]

\[ \begin{aligned} n(e^-)&=2\times0.00600\\\ &=0.0120\ \mathrm{mol} \end{aligned} \]

Therefore, 0.0120 mol of electrons are transferred.

The important point is that the zinc oxidation reaction determines the electron production, while the manganese dioxide reduction reaction consumes exactly the same amount.

Question 2

An alkaline battery contains 0.0300 mol of zinc available for discharge.

It supplies a constant current of 0.180 A for 7.00 hours.

Assume each zinc atom releases two electrons during discharge.

(a) Calculate the theoretical maximum charge available from the zinc.

(b) Calculate the charge transferred during the seven-hour discharge.

(c) Calculate the percentage of the theoretical charge that has been used.

(d) Explain why the actual operating time of a commercial battery may be shorter than the theoretical prediction.

Use:

\[ F=96\\,485\ \mathrm{C\\,mol^{-1}} \]

Solution 2

The theoretical maximum charge is \(5.79\times10^3\) C. The seven-hour discharge uses \(4.54\times10^3\) C, equivalent to approximately 78.4% of the theoretical charge.

(a) Calculate the theoretical maximum charge.

Each zinc atom releases two electrons:

\[ \ce{Zn -> Zn^2+ + 2e^-} \]

Therefore:

\[ n(e^-)=2\times0.0300 \]

\[ n(e^-)=0.0600\ \mathrm{mol} \]

Use Faraday’s law:

\[ Q=n(e^-)F \]

Substitute:

\[ \begin{aligned} Q&=0.0600\times96\\,485\\\ &=5789.1\ \mathrm{C} \end{aligned} \]

Therefore:

\[ Q_{\mathrm{max}}=5.79\times10^3\ \mathrm{C} \]

(b) Calculate the charge transferred in seven hours.

First convert time into seconds:

\[ t=7.00\times3600=25\\,200\ \mathrm{s} \]

Use:

\[ Q=It \]

\[ \begin{aligned} Q&=0.180\times25\\,200\\\ &=4536\ \mathrm{C} \end{aligned} \]

(c) Calculate the percentage of theoretical charge used.

\[ \text{Percentage} = \frac{Q_{\mathrm{used}}}{Q_{\mathrm{max}}} \times100 \]

\[ \begin{aligned} \text{Percentage} &=\frac{4536}{5789.1}\times100\\\ &=78.4\% \end{aligned} \]

(d) Explain why actual operating time may be shorter.

The calculation assumes that every zinc atom reacts completely and that every electron produced contributes to useful external current.

Real batteries do not operate under these ideal conditions.

Some reactants may become inaccessible, internal resistance causes voltage losses, and the device may stop working once the battery voltage falls below its required operating voltage.

Side reactions can also reduce the efficiency of the cell.

Therefore, Faraday’s law provides a theoretical maximum, not a guaranteed operating time.

Question 3

A lithium-ion cell using graphite and lithium cobalt oxide has an open-circuit potential of 3.95 V.

When connected to a device drawing 2.00 A, its measured terminal voltage falls to 3.65 V.

Assume a simplified internal-resistance model.

(a) Calculate the internal resistance of the cell.

(b) Explain why the open-circuit potential and terminal voltage are different.

(c) A student claims that the cell’s standard cell potential must be 3.95 V because this is the voltage measured without a load. Explain why this conclusion is incorrect.

Solution 3

The estimated internal resistance is 0.150 \(\Omega\). Neither measured voltage is automatically the standard cell potential because both depend on the cell’s actual operating conditions.

(a) Calculate internal resistance.

Use:

\[ V_{\mathrm{terminal}}=E_{\mathrm{cell}}-Ir \]

Rearrange:

\[ r=\frac{E_{\mathrm{cell}}-V_{\mathrm{terminal}}}{I} \]

Substitute:

\[ \begin{aligned} r&=\frac{3.95-3.65}{2.00}\\\ &=\frac{0.30}{2.00}\\\ &=0.150\ \Omega \end{aligned} \]

(b) Explain the difference between the voltages.

When the cell is disconnected, essentially no current flows.

Therefore, there is negligible voltage loss across its internal resistance.

Once the cell supplies current, some of its potential difference is lost internally. The terminal voltage consequently decreases.

The greater the current, the greater the voltage drop in this simplified model.

(c) Explain why 3.95 V is not necessarily the standard cell potential.

Standard cell potential, \(E^\circ_{\mathrm{cell}}\), refers to a cell operating under standard conditions.

The measured open-circuit potential of a lithium-ion battery depends on its state of charge and the composition of its electrode materials.

As lithium ions move between graphite and lithium cobalt oxide, the composition of each electrode changes. This affects the cell’s actual potential.

The 3.95 V measurement is therefore an operating-condition value, not proof of the standard cell potential.

The nominal 3.7 V rating is also not a fixed voltage. It represents a typical operating voltage for the cell chemistry.

This question tests an important distinction: standard potential describes a defined electrochemical system under standard conditions, while operating voltage describes what the actual battery supplies.

Question 4

A technology company is choosing between single-use alkaline batteries and rechargeable lithium-ion batteries for a new range of electronic devices.

One engineer argues:

“Rechargeable lithium-ion batteries are always better for the environment because they can be used repeatedly. Single-use alkaline batteries are therefore the less sustainable choice.”

Evaluate this statement.

In your response, compare:

• The redox chemistry and reversibility of each battery.
• The electrode materials and resources required.
• Energy density and practical performance.
• The advantages and limitations of recycling.
• The importance of the intended application.

Solution 4

Rechargeable lithium-ion batteries can reduce material consumption through repeated use, but they are not automatically the more sustainable choice in every application.

The environmental outcome depends on the materials, manufacturing processes, useful lifetime, energy delivered, and end-of-life management.

Redox chemistry and reversibility

Alkaline batteries use zinc oxidation and manganese dioxide reduction to produce electrical energy.

Their electrode materials undergo chemical changes during discharge. Ordinary alkaline cells are not designed to reverse these changes reliably, so the cells are generally discarded once depleted.

Lithium-ion batteries use reversible lithium-ion intercalation between graphite and lithium cobalt oxide electrodes.

During discharge, lithium ions move from graphite towards the cobalt oxide electrode. During charging, an external power supply drives them back towards graphite.

This allows the battery to undergo repeated cycles, reducing the need to manufacture a completely new cell after every discharge.

Materials and resource demand

Alkaline batteries require zinc and manganese dioxide, alongside potassium hydroxide and structural materials.

Lithium cobalt oxide batteries require lithium, cobalt, graphite, copper, aluminium, and organic electrolyte materials.

The demand for cobalt is particularly relevant to LCO batteries. The USGS estimated that the Democratic Republic of the Congo accounted for approximately 76% of global mined cobalt production in 2024.

This concentration creates supply-chain challenges and makes responsible sourcing important.

However, not all lithium-ion batteries use cobalt. LFP batteries, for example, use iron phosphate instead.

Performance

Lithium-ion batteries have higher energy density, making them useful in applications where weight and size matter.

Their ability to recharge also makes them suitable for devices with frequent energy demands.

Alkaline batteries are inexpensive, readily available, and useful in devices with relatively low or intermittent power requirements.

Their single-use design can become a disadvantage when a device needs frequent battery replacement.

Recycling and disposal

Both battery types contain materials that should be managed responsibly.

Lithium-ion batteries offer opportunities to recover valuable materials, but recycling requires suitable collection and processing facilities. Damaged cells also present a fire risk if incorrectly disposed of.

Alkaline batteries contain recoverable metals, including zinc and manganese. However, their environmental advantage depends partly on how effectively they are collected and recycled.

Overall judgement

The engineer’s statement is too broad.

A rechargeable lithium-ion battery is often a sensible choice for a smartphone because it provides high energy density and can be used repeatedly.

An alkaline battery may be a practical option for a low-power household device that requires inexpensive, readily available cells.

The most sustainable option depends on the energy service required and the complete life cycle of the battery, rather than rechargeability alone.

Question 5

Two identical, sealed lithium-ion cells use graphite and lithium cobalt oxide electrodes.

Both cells are labelled with a nominal voltage of 3.7 V.

However, their initial conditions are different:

The cells have identical capacities, are at the same temperature, and have identical voltage-state-of-charge relationships.

They are connected positive-to-positive and negative-to-negative through a current-limited external connection.

Assume that the cells are ideal for the purposes of this question: there are no unwanted side reactions, and no lithium-containing material can cross either cell’s casing. The connection remains in place until the current becomes zero.

(a) Which cell undergoes spontaneous discharge, and which cell is being charged? Explain why the nominal voltage labels do not determine whether current flows.

(b) Describe the direction of electron movement in both external connections.

(c) Describe the direction of lithium-ion movement inside each cell.

(d) Does lithium physically move from cell A into cell B? Explain whether the total lithium content and mass of either sealed cell changes.

(e) Determine the final state of charge of each cell. Explain why the cells reaching the same terminal voltage does not mean that both are fully charged.

(f) Has the standard cell potential of either cell changed during this process? Explain the difference between standard cell potential and the actual cell potential.

Solution 5

Cell A discharges spontaneously and transfers electrical energy to cell B, which charges. At the end of the idealised process, both cells are 50% charged. Neither cell gains or loses lithium overall, and the standard cell potential of their chemistry has not changed.

This problem brings together redox chemistry, electrical potential, ion movement, and conservation of matter.

(a) Identify which cell discharges and which charges.

Cell A initially has an open-circuit voltage of 4.10 V, while cell B has an open-circuit voltage of 3.40 V.

Cell A therefore has the greater initial electrical potential.

It undergoes spontaneous discharge, supplying electrical energy to cell B.

Cell B receives this energy and undergoes charging.

This is an important distinction: the cells have identical nominal voltage ratings, but their actual operating potentials are different.

The 3.7 V labels describe typical operating voltages. They do not guarantee that the cells have the same potential at every state of charge.

The potential difference between the cells provides the driving force for the overall process.

(b) Trace the electron movement.

Electrons move through the external connections.

In the negative connection, electrons travel:

\[ \text{A negative} \longrightarrow \text{B negative} \]

In the positive connection, electrons travel:

\[ \text{B positive} \longrightarrow \text{A positive} \]

Cell A produces electrons through oxidation at its graphite electrode.

Those electrons enter the graphite electrode of cell B, where they are used during the charging reaction.

At the other end, electrons are released from the lithium cobalt oxide electrode of B and return to the positive electrode of A.

The electrons do not pass through either cell’s electrolyte.

(c) Trace lithium-ion movement.

The direction of lithium-ion movement is different in the two cells.

In cell A, which is discharging:

\[ \text{Graphite}\longrightarrow\text{Lithium cobalt oxide} \]

Lithium ions leave the graphite electrode and enter the cobalt oxide electrode.

In cell B, which is charging:

\[ \text{Lithium cobalt oxide}\longrightarrow\text{Graphite} \]

Lithium ions move in the opposite direction.

This is the central feature of lithium-ion battery operation: the ions move internally between the electrodes, while electrons travel through the external connections.

(d) Does lithium move between the cells?

No. Lithium ions do not travel from cell A into cell B.

Each cell is a separate sealed chemical system. Lithium ions move only between the electrodes within their own cell.

Cell A loses lithium from its graphite electrode, but that lithium enters its own cobalt oxide electrode.

Cell B gains lithium in its graphite electrode, but that lithium comes from its own cobalt oxide electrode.

Therefore, neither cell gains nor loses its total lithium inventory.

The same principle applies to the overall mass of each cell. Electrons leave one terminal and enter the other, so there is no net transfer of electrons into or out of either cell. No material crosses either casing.

The chemical composition inside each cell changes, but its total material inventory remains constant.

(e) Determine the final states of charge.

Both cells finish at 50% state of charge.

There are two conditions to consider.

First, the cells are identical and have the same voltage-state-of-charge relationship. At zero current, their terminal potentials must be equal because they are connected in parallel.

Under the stated assumptions, equal potentials correspond to equal states of charge.

Second, the total amount of charge stored across the two cells must be conserved.

Initially:

• Cell A contains 100% of its charge capacity.
• Cell B contains 0% of its charge capacity.

Together, the cells therefore contain one full cell’s worth of stored charge.

Because the cells have identical capacities, an equal final state of charge means that each contains half of this combined amount.

\[ \text{Final state of charge} = \frac{100\%+0\%}{2} \]

\[ \boxed{50\%\text{ for each cell}} \]

This result is particularly important.

Cell B does not become fully charged simply because it receives energy. Cell A also loses stored energy during the transfer.

Some of the energy is dissipated as heat in the external connection, even though the charge-balance argument remains valid.

(f) Has the standard cell potential changed?

No. The standard cell potential is a property of the specified electrochemical reaction under standard conditions.

The actual cell potential, however, changes as the lithium concentration within the electrode materials changes.

During discharge, lithium leaves graphite and enters the cobalt oxide structure. During charging, the reverse occurs.

These changes in electrode composition influence the actual operating voltage.

The two cells eventually reach the same terminal voltage because they have reached the same equilibrium state under the assumptions in the question.

They have not necessarily reached the voltage corresponding to a fully charged cell.

This is the subtle distinction: equal terminal voltages do not mean equal standard potentials, and equal nominal voltage labels do not mean equal initial states of charge.

The energy transfer is possible because the spontaneous discharge of cell A drives the otherwise non-spontaneous charging of cell B.

05Where this leads next

The chemistry of batteries connects several major ideas in HSC electrochemistry.

You can now see how:

• Redox reactions convert chemical energy into electrical energy.
• Standard reduction potentials help predict whether a cell reaction is spontaneous.
• Electrode materials determine how much energy a battery can store and whether its reactions can be reversed.
• Ion movement through electrolytes maintains charge balance.
• Faraday’s law connects the number of electrons transferred to the total electrical charge.
• Cell voltage, capacity, and energy density determine practical battery performance.

The next useful step is to combine reduction-potential calculations with Faraday’s law. This allows you to move beyond identifying which electrode is oxidised or reduced and calculate how much charge and electrical energy a particular redox reaction can actually deliver.

That is the chemistry behind turning a balanced equation into a working battery.