Batteries and Fuel Cells

Every time a torch clicks on, a car starts on a cold morning, or a spacecraft draws power in the vacuum of space, a galvanic cell engineered for practical use is quietly doing electrochemistry's real job — turning a spontaneous redox reaction into usable current. A battery is one or more such cells wired to deliver that current conveniently and reliably; a fuel cell is a special case that never runs out of "battery" as long as fuel keeps flowing in. This section takes the electrode reactions, EMF and cell-notation ideas from earlier electrochemistry and applies them to three named devices you must know at board level: the dry cell (a disposable primary cell), the lead–acid accumulator (a rechargeable secondary cell), and the H₂–O₂ fuel cell (continuous-feed, pollution-free). The skill tested again and again is combining two given electrode half-reactions into one correctly balanced overall reaction — and reading off from it what happens to concentrations, mass, and charge as the device runs.

The core insight

A battery is a galvanic cell (or set of cells) engineered for practical use — primary cells like the dry cell discharge once and are discarded, secondary cells like the lead–acid accumulator can be recharged by driving the discharge half-reactions backward with an external EMF, and fuel cells such as the H₂–O₂ cell generate current continuously by feeding in fuel and oxidant from outside rather than storing them, in every case with the overall cell reaction obtained by adding the anode and cathode half-reactions after their electron counts are matched.

The core idea

Primary vs secondary cells. A primary cell runs one way only — once its reactants are used up, the cell is discarded, because the internal chemistry is not efficiently reversible. A secondary cell can be recharged: passing a current backward through it (itself a small electrolysis) restores the original reactants, so the same physical cell can be discharged and charged many times.

1. Dry cell (Leclanché-type, primary). The zinc container is the anode, oxidised directly: Zn(s)→Zn2+(aq)+2e−\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-. The cathode is a central carbon rod packed in a moist paste of MnO2\text{MnO}_2 and NH4Cl\text{NH}_4\text{Cl}, where manganese dioxide is reduced, MnO2+NH4++e−→MnO(OH)+NH3\text{MnO}_2 + \text{NH}_4^+ + e^- \rightarrow \text{MnO(OH)} + \text{NH}_3. Initial EMF is about 1.5 V1.5\ \text{V}, but it drops steadily on use since the paste composition changes irreversibly — the hallmark of a primary cell.

2. Lead–acid accumulator (secondary). Anode: spongy lead, Pb(s)+SO42−(aq)→PbSO4(s)+2e−\text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^-. Cathode: lead coated in lead dioxide, PbO2(s)+SO42−(aq)+4H+(aq)+2e−→PbSO4(s)+2H2O(l)\text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l). The electrolyte is dilute H2SO4\text{H}_2\text{SO}_4, and since it is a reactant that is consumed on discharge, its concentration (measured as density) is used to gauge the state of charge. Recharging forces both half-reactions in reverse, regenerating Pb\text{Pb}, PbO2\text{PbO}_2 and H2SO4\text{H}_2\text{SO}_4.

Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)\text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)

where: the left side is the discharged-from state (metallic Pb\text{Pb}, PbO2\text{PbO}_2, dissolved acid) and the right side is the discharged-to state (solid PbSO4\text{PbSO}_4 deposited on both plates, water diluting the remaining acid).

3. H₂–O₂ fuel cell (continuous feed). Reactants are supplied from outside rather than stored, so current flows as long as gas is fed in, and the only product is water. In alkaline electrolyte (aqueous KOH) with porous, catalysed (Pt/Pd) electrodes: anode 2H2(g)+4OH−(aq)→4H2O(l)+4e−2\text{H}_2(g) + 4\text{OH}^-(aq) \rightarrow 4\text{H}_2\text{O}(l) + 4e^-, cathode O2(g)+2H2O(l)+4e−→4OH−(aq)\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq), overall 2H2(g)+O2(g)→2H2O(l)2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l). Because there is no combustion step, the cell is not limited by Carnot-type thermal losses, so it is markedly more efficient than burning the same fuel — and pollution-free, which is why fuel cells power spacecraft.

Figure
A schematic galvanic cell showing an anode releasing electrons into an external wire that lights a lamp and returns to a cathode, with ions crossing an electrolyte between the two electrodes
Any battery: oxidation at the anode, reduction at the cathode anode oxidation cathode reduction electrolyte positive ions migrate to cathode lamp electrons flow anode to cathode
Derivation ACombining Half-Reactions With Unequal Electron Counts

The single skill examiners repeatedly test is turning two electrode half-reactions into one overall equation, correctly scaled so that electrons cancel exactly rather than merely approximately, and the dry cell is the sharpest test of it because its anode loses two electrons per zinc atom while its cathode gains only one electron per manganese centre, so the two half-reactions cannot simply be added as written.

Step 1. Write both half-reactions as given, each already balanced in atoms and charge on its own. Anode (oxidation): Zn(s)→Zn2+(aq)+2e−\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-. Cathode (reduction): MnO2+NH4++e−→MnO(OH)+NH3\text{MnO}_2 + \text{NH}_4^+ + e^- \rightarrow \text{MnO(OH)} + \text{NH}_3.

Step 2. Compare the electron counts. The anode releases 2e−2e^-; the cathode consumes only 1e−1e^- per formula unit. Before the two can be added, the number of electrons released must equal the number consumed, so the cathode half-reaction must be scaled up.

Step 3. Find the lowest common multiple of the electron counts (here 22 and 11, so the multiple is 22) and multiply the cathode half-reaction throughout by the factor that makes its electron count match: multiplying by 22 gives 2MnO2+2NH4++2e−→2MnO(OH)+2NH32\text{MnO}_2 + 2\text{NH}_4^+ + 2e^- \rightarrow 2\text{MnO(OH)} + 2\text{NH}_3. The anode is left unscaled since its 2e−2e^- already matches.

Step 4. Add the (now electron-matched) half-reactions and cancel the 2e−2e^- that appears on the product side of the anode and the reactant side of the cathode:

Zn(s)+2MnO2+2NH4+(aq)→Zn2+(aq)+2MnO(OH)+2NH3\text{Zn}(s) + 2\text{MnO}_2 + 2\text{NH}_4^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{MnO(OH)} + 2\text{NH}_3

Step 5. Check the result. Atoms: 11 Zn, 22 Mn, 44 O on the left balance 11 Zn, 22 Mn, 2+2=42+2=4 O on the right (counting the O in MnO(OH)\text{MnO(OH)}); charge: +2+2 on the left (2NH4+2\text{NH}_4^+) balances +2+2 on the right (Zn2+\text{Zn}^{2+}). The same scale-then-add method, with no scaling needed, is exactly what produces the lead–acid and fuel-cell overall equations shown above, where both half-reactions already share 2e−2e^- or 4e−4e^-. Verified: dry cell overall equation atom count Zn 1=1, Mn 2=2, O 4=4, N 2=2, H 8=8, charge +2=+2.

Worked example

Problem: A lead–acid accumulator supplies a steady current until 1.0 F1.0\ \text{F} (one faraday, 96,500 C96{,}500\ \text{C}) of charge has passed through the external circuit. Using the overall discharge reaction Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)\text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l), find how many moles of PbSO4\text{PbSO}_4 are deposited and how many moles of H2SO4\text{H}_2\text{SO}_4 are consumed.

Step 1. Read off the electron count from the electrode half-reactions. Both the anode (Pb→PbSO4+2e−\text{Pb} \rightarrow \text{PbSO}_4 + 2e^-) and the cathode (PbO2→PbSO4\text{PbO}_2 \rightarrow \text{PbSO}_4, gaining 2e−2e^-) involve 22 electrons, matching the overall equation's stoichiometry directly — no separate scaling is needed here, unlike the dry cell.

Step 2. Relate charge passed to moles of electrons. By definition 1 F=96,500 C1\ \text{F} = 96{,}500\ \text{C} is the charge on one mole of electrons, so passing 1.0 F1.0\ \text{F} means 1.0 mol1.0\ \text{mol} of electrons has flowed through the circuit.

Step 3. Use the overall equation's ratio. The balanced equation shows 22 electrons (from the two 2e−2e^- half-reactions, cancelled in combining) produce 2 mol2\ \text{mol} of PbSO4\text{PbSO}_4 and consume 2 mol2\ \text{mol} of H2SO4\text{H}_2\text{SO}_4 — a 1:1:11{:}1{:}1 ratio of electrons to PbSO4\text{PbSO}_4 to H2SO4\text{H}_2\text{SO}_4.

Step 4. Apply the ratio to 1.0 mol1.0\ \text{mol} of electrons: 1.0 mol1.0\ \text{mol} electrons produces 1.0 mol1.0\ \text{mol} PbSO4\text{PbSO}_4 and consumes 1.0 mol1.0\ \text{mol} H2SO4\text{H}_2\text{SO}_4. Answer: 1.0 mol of PbSO4\text{PbSO}_4 is deposited (spread across both plates) and 1.0 mol of H2SO4\text{H}_2\text{SO}_4 is consumed from the electrolyte, which is exactly why the acid's concentration — and hence its measured density — falls as the accumulator discharges.

Common traps

  • Calling the dry cell rechargeable. It is a primary cell; the MnO2\text{MnO}_2/paste cathode chemistry is not efficiently reversible, unlike the lead–acid accumulator.
  • Adding half-reactions without matching electrons first. As with the dry cell (anode 2e−2e^-, cathode 1e−1e^-), the cathode half-reaction must be scaled before adding, or atoms and charge will not balance.
  • Forgetting that recharging a lead–acid battery is itself electrolysis — an external EMF greater than the cell's own forces the discharge half-reactions to run backward.
  • Treating a fuel cell as an energy store. It stores no fuel internally; H2\text{H}_2 and O2\text{O}_2 are fed continuously from external tanks, which is the key difference from primary and secondary cells.

Ask the tutor: curious how many faradays a real car battery actually delivers before recharge, or want to see the H₂–O₂ fuel cell's electrode reactions rewritten for acidic electrolyte instead of alkaline?

Why can't a dry cell be recharged like a lead-acid battery?

Recharging means driving the discharge half-reactions backward with an external EMF, which only works cleanly if the products stay in a form that can be converted back. In the lead-acid cell both electrodes end up as solid PbSO4\text{PbSO}_4 that reforms Pb\text{Pb} and PbO2\text{PbO}_2 on the same spot when current is reversed. In the dry cell, the MnO(OH)\text{MnO(OH)} and NH3\text{NH}_3 formed at the cathode diffuse away and react further in the moist paste, so the original MnO2\text{MnO}_2 cannot be efficiently regenerated — the chemistry is not reversible in practice, which is exactly what defines a primary cell.

Why is a fuel cell more efficient than just burning hydrogen in an engine?

Burning hydrogen to run a heat engine first converts chemical energy to heat, then heat to mechanical work, and that second step is capped by Carnot-type thermal efficiency limits — a large fraction of the energy is unavoidably lost as waste heat. A fuel cell skips the heat step entirely: the H2\text{H}_2–O2\text{O}_2 reaction converts chemical energy directly into electrical energy at the electrodes, so it is not bound by the same thermal ceiling. That is also why its only by-product is water rather than combustion exhaust.

If the electron counts already match in the lead-acid and fuel cell half-reactions, why does the dry cell need scaling at all?

It comes down to how many electrons each named ion or molecule happens to gain or lose in its specific reduction or oxidation step, which is fixed by the reaction's chemistry, not by convenience. Zinc always loses 2 electrons going from Zn\text{Zn} to Zn2+\text{Zn}^{2+}, but the MnO2\text{MnO}_2 reduction step in the dry cell is written as a 1-electron process. Whenever the two half-reactions you are combining don't already share the same electron count, you must scale one or both by whole numbers — using the lowest common multiple — before adding, exactly as shown in the derivation above.

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Where this leads