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Simple Electric Cells

Learn how simple cells produce electrical energy: use the reactivity series to identify polarity, electron flow, observations and half-equations.

  • SEC G3 Pure Chemistry 2027

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A simple cell uses a redox reaction to supply electrical energy to an external circuit. In the two-metal cells with dilute acid studied here, compare the metals to identify the electron supplier, then use the electrolyte to identify what accepts those electrons. The wire and the ionic path through the electrolyte must both be complete for a sustained current through the load.

1. Definition

A. Simple Electric Cell

A simple electric cell converts chemical energy into electrical energy using a redox reaction between two electrodes in an electrolyte.

B. Key Sign Difference

In a simple cell:

  • Anode is negative (oxidation happens here).
  • Cathode is positive (reduction happens here).

2. Key Ideas

A. Follow the two charge pathways

  • Use two suitable different metals, an electrolyte and a complete external circuit.
  • For the acid cells here, the electrolyte supplies hydrogen ions. Do not predict a cathode product from the metal pair alone for every possible electrolyte.
  • The more reactive metal is oxidised more easily, so it becomes the anode (negative).
  • The less reactive metal becomes the cathode (positive).
  • Electrons flow through the wire from anode → cathode.
Zinc and copper simple electric cell in dilute sulfuric acidZinc is the negative anode and copper is the positive cathode. Zinc atoms lose electrons and form zinc ions. Electrons flow through the external circuit from zinc to copper. Hydrogen ions gain electrons at copper and form hydrogen gas. Ions, rather than electrons, carry charge through the electrolyte.Zinc–copper simple cellexternal loadelectrical energy outelectron flow through wiredilute sulfuric acid: ions carry charge through the electrolyteANODE (−)CATHODE (+)ZnCuzinc dissolveshydrogen bubbles formZn²⁺H⁺H⁺H₂(g)Anode: Zn(s) → Zn²⁺(aq) + 2e⁻Cathode: 2H⁺(aq) + 2e⁻ → H₂(g)

Zinc, anode (−): zinc atoms lose electrons and enter solution as zinc ions.

External circuit: electrons travel from zinc, through the load, to copper.

Copper, cathode (+): hydrogen ions gain electrons on copper and form hydrogen gas.

Electrolyte: ions carry charge through the dilute acid. No sustained current passes through the external load if the wire is broken.

In this zinc–copper cell, zinc atoms are oxidised and enter the solution as zinc ions. Electrons travel through the external circuit to copper, where hydrogen ions are reduced to hydrogen gas. The diagram shows the intended external-circuit reactions, not every possible local reaction. Ion colours and sizes are symbolic; other electrolyte ions are omitted. Apparatus and bubbles are schematic and not to scale.
Recall: Reactivity series

The more reactive metal loses electrons more readily. Use the reactivity series before assigning the electrode signs.

Reactivity Series

Electrode sign trap

Electrolysis: anode is positive, cathode is negative. Simple cell: anode is negative, cathode is positive.

What Is Electrolysis?

Syllabus link: hydrogen fuel cell

A hydrogen fuel cell also produces electricity from a redox reaction. Continue to

Hydrogen as a Fuel and Fuel Cells

.

3. Detailed Explanations

Use reactions to identify the electrodes
  • In a simple cell, the anode is negative and the cathode is positive.
  • Oxidation supplies electrons at the anode; reduction accepts them at the cathode.
  • Electrons flow through the external circuit from anode to cathode.
  • Ions carry charge through the electrolyte; electrons do not travel through the liquid between the electrodes.

A. What Happens at Each Electrode

  • Anode (negative): oxidation (electrons are released by reacting particles).
  • Cathode (positive): reduction (electrons are accepted by reacting particles).

B. Why the More Reactive Metal Becomes the Anode

More reactive metals lose electrons more easily, so they oxidise and supply electrons to the circuit.

C. Connect the Three Representations

For the zinc–copper cell in dilute sulfuric acid shown above, the intended external-circuit reactions are:

  • Macroscopic: the zinc electrode becomes smaller and hydrogen bubbles form at copper.
  • Particle level: zinc atoms become Zn²⁺(aq) ions; H + (aq) ions gain electrons and pair to form H₂(g) molecules.
  • Symbolic: the two balanced half-equations show the same electron transfer.

Hydrogen can also form when zinc or magnesium reacts directly with acid at its own surface. Real gas observations depend on the conditions, so do not claim that bubbles can occur only at copper or that seeing bubbles proves current is flowing through the external load. Use the stated reaction and circuit to interpret the observation.

4. Common Mistakes

  • Writing “anode is positive” (that is electrolysis, not a simple cell).
  • Saying electrons move through the electrolyte (wrong: ions move in solution; electrons move in the wire).
  • Assuming the positive copper electrode must gain copper. In the acid cell here, hydrogen ions are reduced on copper; no copper(II) ions have been supplied for copper deposition.
  • Using “electropositive” without explanation: say “more reactive metal loses electrons more easily”.

5. Exam Tips

Must-write phrases

“More reactive metal is oxidised, so it is the anode (negative). Less reactive metal is the cathode (positive).”

  • State electron flow direction: anode → cathode.
  • If asked for observations, state what is seen: gas bubbles at the cathode in an acidic single-electrolyte cell, or a copper coating in a zinc–copper half-cell example.

The signs change between a simple cell and electrolysis, but the reaction definitions do not: anode = oxidation; cathode = reduction. A simple cell supplies electrical energy; electrolysis requires electrical energy from a supply.

6. Worked Examples

Modelled example 1

Single Electrolyte Cell (Zinc + Copper in Acid)

Core

Problem

Zinc and copper are placed in dilute acid and connected by a wire in one electrolyte. Identify both electrodes, state the electron-flow direction, and write the half-equations.
Study the worked solution

  1. Use reactivity to assign the anode

    Method

    Choose zinc as the negative anode.

    Reason

    Zinc is more reactive than copper, so zinc atoms lose electrons more readily.

    Working

    Zn(s) → Zn²⁺(aq) + 2e⁻.

  2. Identify the cathode reaction

    Method

    Reduce hydrogen ions on the positive copper cathode.

    Reason

    Copper provides a conducting surface; hydrogen ions in the acid accept the arriving electrons.

    Working

    2H + (aq) + 2e⁻ → H₂(g).

  3. Connect equations and observations

    Method

    Follow electrons from zinc to copper and combine the half-equations.

    Reason

    The two electrons released at zinc are used at copper.

    Working

    Electron flow: zinc → copper. Overall: Zn(s) + 2H + (aq) → Zn²⁺(aq) + H₂(g). Hydrogen bubbles form on copper and give a pop with a lighted splint.

Common misconception 2

Identify Electrodes (Magnesium + Copper in Dilute Sulfuric Acid)

Find and correct the mistake

Learner claim

Magnesium and copper are placed in dilute H₂SO₄(aq) and connected by a wire. A student says, “Copper must be the negative anode because bubbles appear on it.” Explain the mistake and correct the claim.

Separate electrode identity from observation

Negative anode
Process at copper

View solution step by step

  1. Assign the anode by reactivity

    Method

    Identify magnesium as the negative anode.

    Reason

    Magnesium is more reactive, so it oxidises and supplies electrons.

    Working

    Mg(s) → Mg²⁺(aq) + 2e⁻.

  2. Explain bubbles at copper

    Method

    Identify copper as the positive cathode and hydrogen as the gas.

    Reason

    Hydrogen ions gain the electrons arriving at copper; seeing a product there does not make it the anode.

    Working

    2H + (aq) + 2e⁻ → H₂(g); the gas gives a pop with a lighted splint.

Guided practice 3

Apply the Rules to a New Simple Cell

About 7 min

New metal pair

Iron and copper electrodes are dipped in dilute hydrochloric acid and connected by a wire. Identify both electrodes, state the electron-flow direction and predict one observation at each electrode.

Apply the same reactivity rule

Iron electrode
Copper electrode

Hints

Hint 1: reactivity first
Iron is more reactive than copper, so decide which metal supplies electrons.
Hint 2: ions in the electrolyte
Dilute hydrochloric acid supplies H⁺ ions that can gain electrons.
View solution step by step

  1. Oxidise iron

    Method

    Use iron as the negative electrode.

    Reason

    More-reactive iron atoms lose electrons.

    Working

    Fe(s) → Fe²⁺(aq) + 2e⁻; the iron electrode becomes smaller.

  2. Reduce hydrogen ions

    Method

    Form hydrogen at the positive copper electrode.

    Reason

    H⁺ ions accept the electrons arriving through the wire.

    Working

    2H + (aq) + 2e⁻ → H₂(g); bubbles form on copper.

  3. Combine the transfer

    Method

    State that electrons move from iron to copper.

    Reason

    Iron supplies electrons and hydrogen ions accept them at copper.

    Working

    Fe(s) + 2H + (aq) → Fe²⁺(aq) + H₂(g).

7. Mind Stretchers

Mind stretcher 1: Spot the Sign ErrorExtension

A student writes: “In a simple cell, the anode is positive because oxidation happens there.” Identify the mistake and correct the statement.

Show Answer

Mistake: using electrolysis signs for a simple cell.

Correct: in a simple cell, the anode is negative (oxidation happens there) and the cathode is positive.

Mind stretcher 2: Compare Two CellsExtension

Two cells use the metal pairs Mg/Cu and Zn/Cu. For each cell, identify the negative electrode and state the direction of electron flow.

Show Answer

Magnesium is more reactive than copper, so Mg is the negative electrode in the first cell and electrons flow from Mg to Cu.

Zinc is more reactive than copper, so Zn is the negative electrode in the second cell and electrons flow from Zn to Cu.

Try independently: The wire between zinc and copper in dilute acid is disconnected. A learner sees bubbles at zinc and concludes that the cell must still be delivering electrical energy through the disconnected external load. Explain why the observation does not support that conclusion.

Show answer and reasoning

Zinc can react directly with acid and produce hydrogen at its own surface. With a broken external circuit, there is no sustained electron flow through the external load, even though a chemical reaction can continue locally. Gas formation and electrical energy delivered to the load are different pieces of evidence.

Practise and check

Practise and check

Ready to test your knowledge? Focus on electrode signs, electron flow, and half-equations in simple cells.

Open the Redox Chemistry topic check
Syllabus and review details

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