The battery in your phone is a galvanic cell — a controlled redox reaction where electrons flow through a wire instead of directly between reactants, and the chemistry happening at each electrode determines how long your charge lasts.
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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.
Now imagine separating the zinc and the copper sulfate into two different containers connected by a wire. The same reaction would occur — but electrons would travel through the wire. (1) If electrons are flowing through a wire, what does that mean in terms of electrical current? (2) What problem do you think would arise if the two solutions were not also connected somehow? Write your predictions before reading on.
Type your predictions below — you will revisit them at the end.
Write your predictions in your book. You will revisit them at the end.
📚 Content
A galvanic cell converts chemical energy directly into electrical energy by separating the two halves of a spontaneous redox reaction so that electrons must flow through an external circuit.
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.
| 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.
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.
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:
The lithium-ion (Li-ion) battery in your phone applies every principle from this lesson. During discharge (producing current):
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.
📐 Worked Examples
Cu²⁺/Cu has the more positive E° (+0.34 V) → Cu²⁺ is more readily reduced → copper is the cathode.
Zn²⁺/Zn has the more negative E° (−0.76 V) → Zn is more readily oxidised → zinc is the anode.
Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
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 = E°cathode − E°anode = +0.34 − (−0.76) = +0.34 + 0.76 = +1.10 V
E°cell > 0 → spontaneous. The cell will produce current.
Anode (zinc): zinc metal dissolves → anode decreases in mass; [Zn²⁺] in anode solution increases.
Cathode (copper): Cu²⁺ deposits → cathode increases in mass; [Cu²⁺] in cathode solution decreases.
Fe → Fe²⁺ (oxidation, anode). Zn²⁺ → Zn (reduction, cathode).
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).
Ag → Ag⁺ (oxidation, anode). Cu²⁺ → Cu (reduction, cathode).
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⁺.
✏️ Activities
A student wants to build a galvanic cell using iron (Fe) and silver (Ag) electrodes. Use E°(Fe²⁺/Fe) = −0.44 V and E°(Ag⁺/Ag) = +0.80 V.
Type your working for all five parts.
Show all working in your workbook.
For each cell reaction below, use the standard reduction potential table to calculate E°cell and predict whether the reaction is spontaneous. If spontaneous, identify the anode and cathode.
Type your E°cell calculation and spontaneity prediction for each.
Show your calculations in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: In a galvanic cell, electrons flow through the salt bridge.
Right: 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.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
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)
Type your description for each component.
Write your descriptions in your book.
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)
Type your working for all three parts.
Show all working in your book.
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)
Type your full extended response.
Write your full response in your book.
Go back to your Think First response. Can you now explain precisely why a salt bridge is needed, and what “charge imbalance” means in the context of a galvanic cell?
1. Ag⁺/Ag has more positive E° (+0.80 V) → Ag is cathode. Fe²⁺/Fe has more negative E° (−0.44 V) → Fe is anode.
2. Anode (oxidation): Fe(s) → Fe²⁺(aq) + 2e⁻. Atom: 1Fe ✓. Charge: 0 = +2 − 2 = 0 ✓. Cathode (reduction): Ag⁺(aq) + e⁻ → Ag(s). Atom: 1Ag ✓. Charge: +1 − 1 = 0 ✓.
3. Reduction involves 1e⁻; oxidation involves 2e⁻. Multiply reduction ×2: 2Ag⁺ + 2e⁻ → 2Ag. Overall: Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s). Charge: 0+2=+2; +2+0=+2 ✓.
4. E°cell = +0.80 − (−0.44) = +1.24 V. E°cell > 0 → spontaneous.
5. Iron electrode (anode) gradually dissolves — mass decreases. Silver electrode (cathode) gains mass as silver crystals deposit.
1. Mg = anode (more negative E° = −2.37 V). Cu = cathode (+0.34 V). E°cell = +0.34 − (−2.37) = +2.71 V. Spontaneous. ✓
2. Cu = anode (+0.34 V); Zn²⁺ reduction would require Zn = cathode (−0.76 V). E°cell = −0.76 − (+0.34) = −1.10 V. Non-spontaneous. This reaction does not occur — Cu cannot displace Zn²⁺ (consistent with activity series: Zn > Cu).
Q1 C: Anode: oxidation occurs, metal dissolves, mass decreases, electrons flow OUT (negative terminal). Options A, B, D are all wrong on at least one point.
Q2 B: Fe has more negative E° (−0.44 V) → anode. Ag has more positive E° (+0.80 V) → cathode. E°cell = +0.80 − (−0.44) = +1.24 V.
Q3 B: At cathode: Cu²⁺ ions are reduced to Cu(s) → electrode gains mass; [Cu²⁺] decreases. Option A reverses both. Option D wrongly states concentration increases.
Q4 A: The salt bridge allows ion migration between half-cells to maintain electrical neutrality. Without it, charge imbalance stops the cell. It does NOT carry electrons (that’s the wire).
Q5 C: Al has more negative E° (−1.66 V) → anode. Fe has more positive E° (−0.44 V) → cathode. E°cell = −0.44 − (−1.66) = +1.22 V. Spontaneous.
Q6 D: Without the salt bridge, the anode half-cell accumulates positive charge (Zn²⁺ ions building up) and the cathode half-cell loses positive charge (Cu²⁺ removed). This charge imbalance opposes further electron flow and stops the current.
Q7 A: To maximise E°cell, maximise (E°cathode − E°anode). Use the metal with the most negative E° as anode (Mg, −2.37 V) and the most positive E° as cathode (Ag, +0.80 V). E°cell = +0.80 − (−2.37) = +3.17 V.
Q8: (a) Anode: oxidation occurs; metal dissolves forming cations; mass decreases; [metal ion] in solution increases; negative terminal. (b) Cathode: reduction occurs; metal ions deposit as solid; mass increases; [metal ion] in solution decreases; positive terminal. (c) External wire: conducts electrons from anode to cathode; provides the pathway for current flow. (d) Salt bridge: allows ion migration between the two half-cells to maintain electrical neutrality; without it, charge imbalance stops the cell; anions migrate to anode half-cell, cations to cathode half-cell.
Q9: (a) Mg has more negative E° (−2.37 V) → more readily oxidised → anode. Pb has more positive E° (−0.13 V) → more readily reduced → cathode. (b) Anode (oxidation): Mg(s) → Mg²⁺(aq) + 2e⁻. Cathode (reduction): Pb²⁺(aq) + 2e⁻ → Pb(s). Electrons equal at 2e⁻. (c) E°cell = −0.13 − (−2.37) = +2.24 V. E°cell > 0 → spontaneous.
Q10: (a) Cell voltage = E°cell = E°cathode − E°anode. The higher voltage of a Li-ion battery comes from the larger difference in standard reduction potentials between the lithium/graphite anode (highly negative E°, very strong tendency to be oxidised) and the cobalt oxide cathode (more positive E°). The larger the E° difference, the higher the cell voltage. Zinc has a less negative E° than lithium, giving a smaller difference and lower voltage. (b) A Li-ion battery can be recharged because the electrode reactions (Li⁺ insertion/removal from graphite and LiCoO₂) are reversible — applying an external voltage drives the reactions backwards, restoring the original electrode compositions. A zinc-carbon battery cannot be recharged because the zinc electrode dissolves irreversibly during discharge — the zinc cannot be cleanly reformed at the electrode by applying a reverse current. (c) During discharge, lithium ions stored in the graphite are released: LiC₆ → Li⁺ + C + e⁻ (simplified). The lithium leaving the graphite structure reduces the mass of the anode — a direct result of the oxidation occurring at the anode where matter is being converted from solid electrode to ionic form in solution.
Galvanic Cells