Galvanic Cells
In 1800, Alessandro Volta stacked alternating discs of copper and zinc separated by brine-soaked cloth and produced the world's first steady electrical current — the Volta pile. Napoleon was so impressed he had Volta demonstrate it before the French Academy of Sciences. Every battery you use today traces directly to that stack of metal discs.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
When zinc metal is placed directly into copper sulfate solution, electrons transfer directly from zinc to copper ions — copper deposits on the zinc and heat is released, but no useful electrical work is done.
Key facts
- The four components of a galvanic cell: anode (oxidation, negative), cathode (reduction, positive), external wire (electron flow), salt bridge (ion migration)
- The cell potential formula: E°cell = E°cathode − E°anode; E°cell > 0 means spontaneous
- Electrons flow from anode (−) to cathode (+) through the external circuit; ions migrate through the salt bridge
Concepts
- Why the half-cells must be physically separated — to force electrons through the external wire, producing useful electrical current
- Why the anode is the negative terminal in a galvanic cell — the spontaneous oxidation reaction drives electrons away from the anode into the wire
- Why the salt bridge is essential — to maintain electrical neutrality in each half-cell and allow the reaction to continue
Skills
- Identify anode and cathode from E° values; write electrode half-equations and the overall cell equation
- Calculate E°cell using E°cell = E°cathode − E°anode and state whether the cell is spontaneous
- Describe the observable changes at each electrode (mass and solution concentration) during cell operation
When zinc metal touches copper sulfate solution directly, the reaction happens instantly and violently — all the energy is released as heat. In a galvanic cell, the same reaction is split across two separate containers connected by a wire: the zinc in one half-cell, the copper sulfate in the other. Now the electrons that transfer between zinc and copper must travel through the wire — and that moving charge is an electric current.
When zinc is placed directly into copper sulfate solution, electrons transfer directly and energy is released as heat. A galvanic cell separates the oxidation and reduction half-reactions into two half-cells connected by an external wire and an internal salt bridge. Because electrons cannot jump through solution, they are forced to travel through the wire — this electron flow is electrical current.
The galvanic cell was first demonstrated by Alessandro Volta in 1800 using stacked discs of zinc and copper separated by brine-soaked cloth — the voltaic pile. The fundamental principle is identical in every battery: a spontaneous redox reaction harnessed to do electrical work by forcing electrons through an external circuit.
A galvanic cell converts chemical energy to electrical energy by separating a spontaneous redox reaction into two half-cells. Electrons travel through the external wire (electrical current); ions travel through the salt bridge to maintain neutrality. Both connections are required — remove either and the cell stops.
Pause — copy the highlighted definition into your book before moving on.
Fill the gap: A galvanic cell forces electrons to flow through an external wire by physically separating the two halves of a spontaneous redox reaction into two [___] connected by a salt bridge.
We just saw that a galvanic cell separates a redox reaction into two half-cells connected by a wire and salt bridge. That raises a question: what are the specific roles of each component — anode, cathode, wire, and salt bridge — and how do we remember which does what? This card answers it → AN OX / RED CAT: anode = oxidation (−); cathode = reduction (+).
| Component | Function | Change During Operation |
|---|---|---|
| Anode | Oxidation — metal dissolves into solution as ions | Decreases in mass; anode solution [metal ion] increases |
| Cathode | Reduction — metal ions deposit as solid metal | Increases in mass; cathode solution [metal ion] decreases |
| External wire | Electron flow from anode to cathode | Carries electrical current |
| Salt bridge | Ion migration to maintain electrical neutrality in both half-cells | Ions gradually depleted; anions migrate into anode half-cell, cations into cathode half-cell |
The salt bridge is essential. Without it, charge builds up: the anode half-cell becomes increasingly positive (metal ions accumulating) and the cathode half-cell becomes increasingly negative (metal ions being removed). This charge imbalance stops the cell immediately. The salt bridge (typically KNO₃ or KCl in agar gel) allows ions to migrate, maintaining neutrality and allowing the cell to continue operating.
AN OX / RED CAT: anode = oxidation (−); cathode = reduction (+). Electrons flow anode → cathode through the wire. In the salt bridge, anions migrate to the anode half-cell and cations migrate to the cathode half-cell to maintain electrical neutrality.
Add the highlighted mnemonic and rule to your notes before the check below.
Match it: Match each galvanic cell component to its correct role.
- Anode
- Cathode
- Salt bridge
- External wire
- Carries electron flow from anode to cathode
- Allows ion migration to maintain electrical neutrality
- Reduction occurs here; electrode gains mass
- Oxidation occurs here; electrode loses mass
We just saw the roles of anode, cathode, wire, and salt bridge. That raises a question: how do we predict which electrode becomes the anode and calculate the voltage the cell will produce? This card answers it → standard reduction potentials (E°) quantify each half-reaction's tendency to be reduced; E°cell = E°cathode − E°anode.
The standard reduction potential (E°) measures the tendency of a species to be reduced under standard conditions (25°C, 1 mol/L, 1 atm), relative to the standard hydrogen electrode (SHE = 0.00 V). A more positive E° means a stronger tendency to be reduced — the species is a better oxidant.
| Half-Reaction | E° (V) |
|---|---|
| K⁺(aq) + e⁻ → K(s) | −2.94 |
| Mg²⁺(aq) + 2e⁻ → Mg(s) | −2.37 |
| Al³⁺(aq) + 3e⁻ → Al(s) | −1.66 |
| Zn²⁺(aq) + 2e⁻ → Zn(s) | −0.76 |
| Fe²⁺(aq) + 2e⁻ → Fe(s) | −0.44 |
| Pb²⁺(aq) + 2e⁻ → Pb(s) | −0.13 |
| 2H⁺(aq) + 2e⁻ → H₂(g) | 0.00 |
| Cu²⁺(aq) + 2e⁻ → Cu(s) | +0.34 |
| Ag⁺(aq) + e⁻ → Ag(s) | +0.80 |
| Au³⁺(aq) + 3e⁻ → Au(s) | +1.50 |
Cell potential formula: E°cell = E°cathode − E°anode
The cathode has the more positive (or less negative) E° value. If E°cell > 0 → spontaneous. If E°cell < 0 → non-spontaneous.
E°cell = E°cathode − E°anode. Both values are taken from the reduction table — do NOT reverse the anode sign before subtracting. More positive E° → more easily reduced → cathode. If E°cell > 0 → spontaneous. Zn/Cu example: +0.34 − (−0.76) = +1.10 V.
Pause — write the highlighted formula and example into your book.
Odd one out: Which of these is NOT a true statement about E°cell = E°cathode − E°anode?
We just saw how to calculate E°cell from standard reduction potentials. That raises a question: as the cell runs over time, what physically changes — in the electrodes, solutions, and voltage? This card answers it → anode loses mass, anode [ion] increases; cathode gains mass, cathode [ion] decreases; cell voltage falls toward zero.
A galvanic cell does not operate in a static state — as the reaction proceeds, all components change in measurable ways. For a zinc-copper galvanic cell:
- Anode (zinc): zinc metal oxidises and dissolves (Zn → Zn²⁺ + 2e⁻). Anode mass decreases. [Zn²⁺] in anode solution increases.
- Cathode (copper): Cu²⁺ ions are reduced and deposit as solid copper. Cathode mass increases. [Cu²⁺] in cathode solution decreases.
- Salt bridge: anions (e.g. NO₃⁻) migrate into the anode half-cell to balance the accumulating Zn²⁺ cations. Cations (e.g. K⁺) migrate into the cathode half-cell to replace the removed Cu²⁺.
- Cell voltage: decreases over time as concentrations approach equilibrium. When E°cell ≈ 0, the cell is flat.
During Zn/Cu cell operation: anode (Zn) mass decreases; [Zn²⁺] in anode solution increases. Cathode (Cu) mass increases; [Cu²⁺] in cathode solution decreases. Cell voltage gradually decreases as concentrations approach equilibrium; when E°cell ≈ 0, the cell is flat.
Add the highlighted changes to your notes before the check below.
Fill the gap: During operation of a Zn/Cu galvanic cell, the mass of the copper cathode [___] as Cu²⁺ ions are reduced and deposited as solid copper.
We just saw the changes that occur as a galvanic cell runs down. That raises a question: how does this same galvanic cell chemistry work in rechargeable lithium-ion batteries — and why do they produce ~3.7 V rather than 1.1 V? This card answers it → Li has a very negative E° value, so the potential difference with the cobalt oxide cathode is much larger than in a Zn/Cu cell.
The lithium-ion (Li-ion) battery in your phone applies every principle from this lesson. During discharge (producing current):
- Anode (graphite): Lithium ions stored in graphite are released: Li⁺(aq) + e⁻ are produced (oxidation). Electrons flow through the external circuit.
- Cathode (lithium cobalt oxide, LiCoO₂): Li⁺ ions and electrons are accepted: Li⁺ + CoO₂ + e⁻ → LiCoO₂ (reduction). Mass of cathode increases.
The cell voltage is approximately 3.6–3.7 V — higher than a standard zinc-copper cell (1.10 V) because the reduction potential difference between the lithium anode and the cobalt oxide cathode is larger. Li-ion cells are rechargeable because the redox reaction is reversible — applying an external voltage drives the reaction in reverse, returning lithium ions to the graphite anode.
Li-ion battery (discharge): graphite anode releases Li⁺ (oxidation); LiCoO₂ cathode accepts Li⁺ and e⁻ (reduction). Voltage ~3.7 V because Li has a very negative E°. Rechargeable because the reaction is reversible — applying external voltage returns Li⁺ to graphite.
Pause — write the highlighted point into your book.
Match it: Match each Li-ion battery component or process to the correct description during discharge.
- Graphite anode
- LiCoO₂ cathode
- ~3.7 V cell voltage
- Rechargeability
- External voltage reverses the reaction, returning Li⁺ to graphite
- Result of the large E° difference between Li anode and cobalt oxide cathode
- Accepts Li⁺ and electrons; increases in mass (reduction)
- Loses Li⁺ ions and releases electrons (oxidation)
Worked examples · reveal as you go
Problem: A galvanic cell uses a zinc electrode in zinc sulfate solution and a copper electrode in copper sulfate solution, connected by a salt bridge and external wire. Use E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V. (a) Identify the anode and cathode. (b) Write the half-equations and overall cell equation. (c) Calculate E°cell. (d) Describe one change at each electrode during operation.
Zn²⁺/Zn has the more negative E° (−0.76 V) → Zn is more readily oxidised → zinc is the anode.
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)
Electrons equal (2e⁻) — add directly and cancel electrons:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Charge check: left = 0 + (+2) = +2. Right = +2 + 0 = +2. ✓
E°cell > 0 → spontaneous. The cell will produce current.
Cathode (copper): Cu²⁺ deposits → cathode increases in mass; [Cu²⁺] in cathode solution decreases.
Problem: Predict whether each reaction is spontaneous under standard conditions. Show working using E°cell = E°cathode − E°anode. (a) Fe(s) + Zn²⁺(aq) → Fe²⁺(aq) + Zn(s). (b) Ag(s) + Cu²⁺(aq) → Ag⁺(aq) + Cu(s).
E°anode (Fe²⁺/Fe) = −0.44 V. E°cathode (Zn²⁺/Zn) = −0.76 V.
E°cell = −0.76 − (−0.44) = −0.76 + 0.44 = −0.32 V
E°cell < 0 → non-spontaneous. Fe cannot displace Zn²⁺ from solution (consistent with activity series: Zn > Fe).
E°anode (Ag⁺/Ag) = +0.80 V. E°cathode (Cu²⁺/Cu) = +0.34 V.
E°cell = +0.34 − (+0.80) = −0.46 V
E°cell < 0 → non-spontaneous. Silver does not displace copper from solution (Ag is below Cu in the activity series).
(a) E°cell = −0.32 V → non-spontaneous. Fe cannot displace Zn²⁺.
(b) E°cell = −0.46 V → non-spontaneous. Ag cannot displace Cu²⁺.
Key Formulas — This Lesson
Common errors · the 3 traps that cost marks
Common misconception
In a galvanic cell, electrons flow through the salt bridge.
Fix: Electrons flow through the external wire from anode to cathode. The salt bridge allows ions to flow internally to maintain charge balance — it does not conduct electrons. Confusing electron flow with ion migration is a common source of lost marks.
The anode is always the positive electrode
A student labels the anode as "+" in a galvanic cell diagram, confusing it with an electrolytic cell.
Fix: In a galvanic cell, the anode is the negative terminal (−). The cathode is the positive terminal (+). The confusion arises because in both galvanic and electrolytic cells the anode is where oxidation occurs — but the polarity is opposite. Remember: in a galvanic cell the spontaneous oxidation at the anode pushes electrons out of the anode, making it negative. "AN OX / RED CAT" tells you where reactions occur — not the terminal sign.
Salt bridge ions flow in the wrong direction
A student states that cations from the salt bridge migrate into the anode half-cell to balance the build-up of positive metal ions.
Fix: Anions (e.g. NO₃⁻) migrate from the salt bridge into the anode half-cell to balance the accumulating positive metal ions (Mⁿ⁺). Cations (e.g. K⁺) migrate into the cathode half-cell to replace the positive ions being removed from solution. Ion migration is driven by the need to maintain electrical neutrality — anions go where cations accumulate, cations go where cations are removed.
Quick-fire practice · 5 reps +2 XP per reveal
Q8 (4 marks): Describe the role of each component of a galvanic cell: (a) the anode, (b) the cathode, (c) the external wire, and (d) the salt bridge. Include what happens to the electrode mass and solution concentration for (a) and (b).
Q9 (4 marks): A galvanic cell is constructed using a magnesium electrode in magnesium sulfate solution and a lead electrode in lead(II) nitrate solution. Use E°(Mg²⁺/Mg) = −2.37 V and E°(Pb²⁺/Pb) = −0.13 V. (a) Identify the anode and cathode with justification. (b) Write the balanced half-equation at each electrode. (c) Calculate E°cell and state whether the cell is spontaneous.
Q10 (5 marks): A lithium-ion battery in a phone has a cell voltage of 3.7 V. A zinc-carbon AA battery has a cell voltage of 1.5 V. (a) Explain, using the concept of standard reduction potentials, why the lithium-ion battery has a higher voltage than the zinc-carbon battery. (b) Explain why a lithium-ion battery can be recharged but a standard zinc-carbon battery cannot, using the concept of reaction reversibility. (c) When a lithium-ion battery is being discharged, the graphite anode gradually loses mass. Explain this observation using the relevant half-equation.
A Mg/Pb galvanic cell uses E°(Mg²⁺/Mg) = −2.37 V and E°(Pb²⁺/Pb) = −0.13 V. Identify the anode and cathode, then calculate E°cell.
For the Mg/Pb galvanic cell above, describe one observable change at each electrode during operation.
Alessandro Volta's 1800 copper-zinc pile worked because copper and zinc have different reduction potentials — Zn is more easily oxidised (Zn → Zn²⁺ + 2e⁻, E° = −0.76 V) while Cu²⁺ is more easily reduced (Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V). Electrons flow spontaneously from the zinc (anode) through the external circuit to the copper (cathode). The brine-soaked cloth acted as a salt bridge — allowing ion flow to maintain charge balance without the half-cells mixing. Without it, charge would accumulate and stop the reaction. The calculated cell potential: E°cell = +0.34 − (−0.76) = +1.10 V — which is why a zinc-copper cell produces about 1 volt.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q1. 8. (4 marks) Describe the role of each component of a galvanic cell: (a) the anode, (b) the cathode, (c) the external wire, and (d) the salt bridge. Include what happens to the electrode mass and solution concentration for (a) and (b). (1 mark each)
Q2. 9. (4 marks) A galvanic cell is constructed using a magnesium electrode in magnesium sulfate solution and a lead electrode in lead(II) nitrate solution. Use E°(Mg²⁺/Mg) = −2.37 V and E°(Pb²⁺/Pb) = −0.13 V. (a) Identify the anode and cathode with justification. (b) Write the balanced half-equation at each electrode. (c) Calculate E°cell and state whether the cell is spontaneous. (1 + 2 + 1 marks)
Q3. 10. (5 marks) A lithium-ion battery in a phone has a cell voltage of 3.7 V. A zinc-carbon AA battery has a cell voltage of 1.5 V. (a) Explain, using the concept of standard reduction potentials, why the lithium-ion battery has a higher voltage than the zinc-carbon battery. (b) Explain why a lithium-ion battery can be recharged but a standard zinc-carbon battery cannot, using the concept of reaction reversibility. (c) When a lithium-ion battery is being discharged, the graphite anode gradually loses mass. Explain this observation using the relevant half-equation. (2 + 2 + 1 marks)
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