Batteries — primary (dry cell), secondary (lead-acid, Li-ion): Question bank

True or false — justify

The anode is always the negative terminal of a battery.

In a discharging battery yes — oxidation happens there and pushes electrons out into the wire, making it negative. But during charging the same physical plate is forced to accept electrons (reduction), so "anode" (site of oxidation) actually shifts to the other plate. The label "anode" tracks the reaction, not a fixed wire.

Electrons flow through the electrolyte inside a battery.

False. Electrons travel only through the external wire; inside, the circuit is completed by ions drifting through the electrolyte. If electrons could cross the electrolyte directly, that's just an internal short and no useful work is extracted.

A primary cell can never be recharged, no matter what voltage you apply.

Practically true, but the real reason is that the discharge products are consumed/dispersed or form phases that don't reverse cleanly (e.g. NH₃ escapes, MnO₂→Mn₂O₃ isn't a tidy solid-state reversal). Forcing current mostly electrolyses water and builds pressure — dangerous, not a recharge.

During lead-acid discharge, the electrolyte gets more concentrated.

False — it gets more dilute. Discharge consumes H₂SO₄ and produces H₂O, so density drops from ~1.25 to ~1.15 g/cm³. That falling density is exactly what a hydrometer reads as "low charge".

In a lead-acid cell, only the cathode ends up coated in PbSO₄.

False. Both electrodes convert to PbSO₄ on discharge: Pb (anode) → PbSO₄ and PbO₂ (cathode) → PbSO₄. That's why the reaction is nicknamed the "double-sulfate" reaction.

A higher standard electrode potential means that half-reaction wants to be oxidised.

False — it means it wants to be reduced (gain electrons). See 2.7.05-Standard-electrode-potentials: the more positive the reduction potential, the stronger the pull for electrons, so it becomes the cathode.

Zinc oxidising in a dry cell releases energy because Zn is "reactive".

Loosely true, made precise: Zn²⁺/Zn has a reduction potential of −0.76 V, so Zn loses electrons readily. Coupled to a high-potential cathode, the overall

Δ G < 0

(see 3.1.12-Gibs-free-energy-and-spontaneity), which is what makes the cell spontaneous.

Li-ion batteries store energy by dissolving lithium metal at the anode.

False. They use intercalation — Li⁺ ions slot into and out of layered host structures (graphite, metal oxides) without plating out as metal. Actual Li-metal plating is a failure mode that causes dangerous dendrites.

Spot the error

E_{cell} = E_{anode} - E_{cathode}

, so we subtract cathode from anode."

Wrong order. The correct formula is

E_{cell} = E_{cathode} - E_{anode}

(both as reduction potentials). Flipping it gives a negative EMF and wrongly implies a spontaneous cell is non-spontaneous.

"The dry cell gives 2.0 V because

+ 1.2 - (- 0.76) \approx 2.0

, and that's what you measure."

The 2.0 V is only the theoretical value. Real dry cells read ~1.5 V because of the paste electrolyte, ion-transport resistance, and overpotential losses that the ideal formula ignores.

"Since current is 0.3 A, exactly 0.3 electrons flow per second."

A confusion of units. 0.3 A means 0.3 coulombs per second; dividing by

e = 1.6 \times 1 0^{- 19}

C gives ≈

1.9 \times 1 0^{18}

electrons per second. Amperes count charge, not particles.

"Each Zn atom gives 1 electron, so moles of Zn equal moles of electrons."

Error in stoichiometry.

Zn \to Zn^{2 +} + 2 e^{-}

releases two electrons per atom, so moles of Zn = (moles of

e^{-}

)/2. Skipping the factor of 2 doubles your predicted metal consumption.

"In lead-acid charging, we apply exactly 2.0 V to reverse the reaction."

You must apply more than the cell EMF (>2.0 V per cell) to overcome the reverse reaction plus internal resistance and overpotential. Exactly 2.0 V leaves the system at equilibrium — no net charging happens.

"MnO₂ is the anode of the dry cell because it's the reactive material."

MnO₂ is the cathode (where reduction, MnO₂→Mn₂O₃, occurs). The zinc can is the anode. Being chemically active doesn't decide the role — being reduced vs oxidised does.

Q = I t

gave 600 C = 600 mol of electrons for the car-starter problem."

Coulombs are not moles. Divide by Faraday's constant:

600/96485 \approx 6.2 \times 1 0^{- 3}

mol of electrons. Treating coulombs as moles overestimates by a factor of ~96 000.

Why questions

Why do we physically separate oxidation and reduction instead of mixing the reactants?

If mixed, electrons hop directly between reactants and the energy dumps out as heat, doing no useful work. Separating the half-reactions forces electrons through the external wire, so we harvest the energy as electricity — see 2.7.04-Galvanic-cells-and-cell-notation.

Why does the dry cell form the complex

[Zn(NH_{3})_{2}] Cl_{2}

To remove built-up Zn²⁺ near the anode. By Le Chatelier's principle, accumulating product would slow the oxidation; precipitating Zn²⁺ into an insoluble complex keeps the reaction driving forward.

Why is PbSO₄ chosen (well, formed) rather than a soluble lead product in lead-acid cells?

PbSO₄ is sparingly soluble, so it deposits on the electrode surface. That keeps the lead in place and electrically connected, allowing it to be converted back to Pb / PbO₂ during charging — essential for reversibility.

Why does a battery's voltage drop as it discharges even before it "dies"?

As reactants deplete, the ratio of products to reactants rises, and by the Nernst equation the cell potential falls below its standard value. Concentration, not just "running out", pushes the voltage down.

Why can secondary batteries be recharged but primary ones essentially cannot?

Secondary-cell products (PbSO₄, intercalated Li⁺) stay put as reversible solid phases that external voltage can convert back. Primary-cell products disperse, form gases (NH₃), or undergo non-reversible solid changes, so there's nothing tidy to reverse.

Why does the lead-acid EMF come out to 2.0 V per cell, yet car batteries read 12 V?

One cell gives

1.69 - 0.36 = 2.0

V. Six such cells wired in series add their voltages:

6 \times 2.0 = 12

V. Series stacking sums EMFs.

Why is a battery just a "portable spontaneous redox reaction"?

Because the overall cell reaction has

Δ G < 0

(spontaneous, 3.1.12-Gibs-free-energy-and-spontaneity); the cell merely routes that spontaneous electron transfer through a wire. No spontaneity, no self-driven current.

Edge cases

What happens to a lead-acid battery left fully discharged for a long time?

The fine PbSO₄ crystals grow into large, hard, insulating crystals ("sulfation"), which resist reconversion during charging. The cell loses capacity permanently — a boundary case where "reversible" chemistry stops being reversible.

If you keep discharging a dry cell to 0.9 V and beyond, what's happening chemically?

The active MnO₂ and Zn are nearly exhausted and product concentrations are high, so by Nernst the EMF collapses. Pushing further can rupture the zinc can and leak corrosive paste — the degenerate "dead cell" regime.

At absolute zero of charge (a hypothetical "empty" cell), what is the EMF?

When reactants are fully consumed the reaction quotient blows up, driving the Nernst-predicted EMF toward its lower limit (effectively ~0 net useful voltage). No reactant gradient means no driving force.

What if you connect a battery's terminals directly with a thick wire (short circuit)?

The external resistance drops near zero, so current spikes to a huge value limited only by internal resistance. Enormous

I^{2} R

heating results — the cell overheats, may vent, or in Li-ion cells thermally runs away. This is the zero-load limiting case.

Can a "cathode" ever be the negative terminal?

Yes — during charging of a secondary cell, the electrode undergoing reduction (the cathode) is driven by the charger and is connected to the charger's negative terminal. Terminal sign depends on whether the cell is delivering or receiving current.

What limits how thin a car battery's momentary current draw can safely last?

Not the total material (a 3-second, 200 A starter pulse consumes <0.1% of a cell's PbO₂) but heat and gas build-up. The stoichiometric consumption is trivial; the practical limit is thermal, not chemical, in short bursts.

Does corrosion of a metal share the same mechanism as a battery?

Yes — corrosion is a short-circuited galvanic cell on a single metal surface, with anodic and cathodic patches. See 2.7.09-Corosion-and-prevention; the "battery" you don't want is rust forming spontaneously.

Recall Quick self-test

Discharge of lead-acid: acid gets more ___ and water is ___. ::: dilute; produced Dry-cell anode material and its role. ::: Zinc can; it is oxidised (Zn → Zn²⁺ + 2e⁻) Correct EMF formula. ::: $E_{cell} = E_{cathode} - E_{anode}$ (reduction potentials) Li-ion storage mechanism in one word. ::: Intercalation Why primary cells don't recharge, in one phrase. ::: Discharge products are not cleanly reversible