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SAChemistrySyllabus dot point

How do primary and secondary cells store and supply electrical energy?

Distinguish primary and secondary cells, describe their electrode reactions, and evaluate them as energy resources.

The difference between primary and secondary (rechargeable) cells, their electrode half-equations, the lead-acid and lithium-ion examples, and worked SACE-style charge-capacity and electrode calculations, with a sustainability evaluation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Lead worked calculation
Primary versus secondary cells
Electrode reactions
Common examples
Evaluating batteries as energy resources
Why it matters for managing resources

What this dot point is asking

SACE expects you to distinguish primary and secondary cells, write or interpret their electrode half-equations, perform charge calculations, and evaluate them as energy resources.

Lead worked calculation

Charge delivered and mass of reactant

A primary cell delivers a steady $0.250\ \text{A}$ for $40.0\ \text{hours}$ . Its anode reaction is $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ . Find the mass of zinc consumed. ( $F = 96500\ \text{C mol}^{-1}$ ; $M(\text{Zn}) = 65.4\ \text{g mol}^{-1}$ .)

Step 1: Calculate the total electrical charge delivered

Current is defined as charge per unit time, so rearranging gives $Q = It$ . The key trap here is that $I$ is in amperes (coulombs per second), so time must be in seconds. Convert hours to seconds by multiplying by $3600$ before substituting. Skipping this conversion is the most common arithmetic error in these questions.

Q = It = 0.250 \times (40.0 \times 3600) = 0.250 \times 1.44 \times 10^{5} = 3.60 \times 10^{4}\ \text{C}

Step 2: Convert total charge to moles of electrons

The Faraday constant $F = 96500\ \text{C mol}^{-1}$ is the charge carried by one mole of electrons. Dividing the total charge by $F$ therefore tells you how many moles of electrons flowed through the external circuit during the $40.0$ hours of operation. This is the bridge between the electrical measurement and the chemistry happening inside the cell.

n(e^-) = \frac{Q}{F} = \frac{3.60 \times 10^{4}}{96500} = 0.373\ \text{mol}

Step 3: Use the half-equation to find moles of zinc consumed

The anode half-equation $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ shows that each zinc atom releases exactly two electrons when it is oxidised. This stoichiometric ratio links moles of electrons to moles of zinc: if you know how many moles of electrons were produced, you simply divide by two to find how many moles of zinc atoms were consumed.

n(\text{Zn}) = \frac{n(e^-)}{2} = \frac{0.373}{2} = 0.1865\ \text{mol}

Step 4: Convert moles of zinc to mass

Now that you have the amount of zinc in moles, multiply by the molar mass to convert to the mass that can be measured on a balance. This is the reverse of the $n = m/M$ conversion: $m = nM$ .

m(\text{Zn}) = nM = 0.1865 \times 65.4 = 12.2\ \text{g}

Final answer: $12.2\ \text{g}$ of zinc is consumed. The same logical chain, $Q = It$ , then divide by $F$ , then divide by the electrons-per-atom ratio from the half-equation, applies to any electrochemical calculation.

Primary versus secondary cells

The key requirement for a secondary cell is that the discharge products stay where they form (on or near the electrodes) so they can be converted back. In the lead-acid cell, both electrodes become $\text{PbSO}_4$ , which recharging converts back to $\text{Pb}$ and $\text{PbO}_2$ .

Electrode reactions

During discharge every battery acts as a galvanic cell:

The anode (negative on discharge) is oxidised, releasing electrons.
The cathode (positive on discharge) is reduced, accepting electrons.

During recharge a secondary cell acts as an electrolytic cell: an external supply forces the reactions in reverse, so the electrode that was the anode now undergoes reduction. This reversal is exactly why secondary cells link the galvanic and electrolytic ideas.

Common examples

Evaluating batteries as energy resources

A balanced evaluation weighs several factors:

Rechargeability: secondary cells reduce waste and long-term cost; primary cells are simpler and hold charge longer when idle.
Energy density: lithium-ion stores more energy per unit mass, important for portable and vehicle use.
Cost: secondary cells cost more upfront but less per use over their life.
Lifespan: secondary cells lose capacity over repeated cycles.
Environmental impact: extraction of lithium, cobalt and lead, and safe disposal or recycling of toxic and reactive materials, are significant sustainability concerns.

Why it matters for managing resources

Batteries store energy for portable devices, vehicles and renewable-power grids, so understanding primary versus secondary chemistry, and evaluating their cost, lifespan and environmental footprint, is central to managing energy resources sustainably as society electrifies.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20215 marksA lead-acid car battery uses the discharge reactions: anode

\text{Pb} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 + 2e^-

and cathode

\text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O}

. (a) Classify the cell as primary or secondary and justify. (b) Calculate the mass of lead consumed at the anode when the battery delivers

2.00\ \text{A}

for

3.00\ \text{hours}

. (

F = 96500\ \text{C mol}^{-1}

;

M(\text{Pb}) = 207.2\ \text{g mol}^{-1}

Show worked answer →

(a) It is a secondary (rechargeable) cell: the discharge products ( $\text{PbSO}_4$ ) remain on the electrodes and the reaction can be reversed by applying an external voltage to recharge it. (1 mark)

(b) $Q = It = 2.00 \times (3.00 \times 3600) = 2.00 \times 10800 = 2.16 \times 10^{4}\ \text{C}$ . (1 mark)

$n(e^-) = \dfrac{2.16 \times 10^{4}}{96500} = 0.2238\ \text{mol}$ . The anode uses $2e^-$ per $\text{Pb}$ , so $n(\text{Pb}) = \dfrac{0.2238}{2} = 0.1119\ \text{mol}$ . (2 marks)

$m(\text{Pb}) = nM = 0.1119 \times 207.2 = 23.2\ \text{g}$ . (1 mark)

SACE 20194 marksCompare a primary alkaline cell with a secondary lithium-ion cell. (a) State the key chemical difference between primary and secondary cells. (b) Give one advantage and one disadvantage of secondary cells as an energy resource compared with primary cells.

Show worked answer →

(a) In a primary cell the redox reaction is not readily reversible, so once the reactants are used up the cell is discarded. In a secondary cell the reaction can be reversed by an external voltage, so the cell can be recharged and reused many times. (2 marks)

(b) Advantage: secondary cells can be recharged and reused, reducing waste and long-term cost. Disadvantage: they have a higher initial cost, gradually lose capacity over many cycles, and may contain reactive or toxic materials needing careful disposal. (2 marks)