Batteries — primary (dry cell), secondary (lead-acid, Li-ion): Worked examples

This page is the problem-solving companion to the parent battery note. There we learned what the reactions are. Here we drill the arithmetic — the same four tools appear again and again, so let us name them once and then hit every kind of question an exam can ask.

Intuition The four tools you will keep reaching for

Almost every battery calculation is one (or a chain) of these:

Charge from current & time (tool 1): $Q = I \cdot t$ — "how much electric charge flowed?"
Faraday's bridge (tool 2): $Q = n_{e^{-}} \cdot F$ where $F = 96485 C/mol$ — "how many moles of electrons is that charge?"
Stoichiometry (tool 3): the balanced half-reaction tells us how many electrons ride on each atom or molecule, so we scale from moles of electrons to moles of substance.
Molar-mass conversion (tool 4): $m = n \cdot M$ — "turn moles of substance into grams we can weigh." We call this tool 4 (the "M" step) so it has a name of its own; it is the very last arrow on the map below.

If you can chain current → charge → moles of electrons → moles of substance → mass, you can solve nearly everything below.

Before any tool, we recall the vocabulary from 2.7.01-Oxidation-numbers-and-redox-definition and 2.7.04-Galvanic-cells-and-cell-notation: the anode is where oxidation happens (electrons leave), the cathode is where reduction happens (electrons arrive), and the cell EMF is the voltage difference from 2.7.05-Standard-electrode-potentials.

The scenario matrix

Every question this topic can throw is one of these case classes. Each worked example below is tagged with the cell it covers; jump straight to any row using the section links in the rightmost column.

Cell	Case class	What makes it tricky	Example (jump to)
A	Basic charge → mass (primary cell)	plain tool-chain, 2 electrons per atom	Ex 1
B	EMF from electrode potentials, sign handling	subtract a negative potential correctly	Ex 2
C	Both electrodes consumed (lead-acid)	mass at two electrodes at once	Ex 3
D	Reverse direction (charging)	reaction runs backward, quantities regenerate	Ex 4
E	Limiting-reactant / battery lifetime	when does a reactant hit zero?	Ex 5
F	Concentration change and EMF drift (Nernst)	acid dilutes → voltage sags	Ex 6
G	Real-world word problem	translate words → numbers	Ex 7
H	Exam twist: 1-electron ion (Li⁺) & capacity in mAh	different $z$ , unfamiliar unit	Ex 8
I	Degenerate / sanity edge case	zero current, or 100% depleted	Ex 9

We now walk each cell in order.

Setting the constants once

How to read the tool-chain map

Figure — Batteries — primary (dry cell), secondary (lead-acid, Li-ion)

Alt text / full description (so you can reconstruct the picture without seeing it): On a blueprint-blue gridded background there are five labelled boxes. Four of them carry a step number badge (1, 2, 3, 4) in the corner so the sequence is readable without relying on colour, and each box has a distinct outline style: the two "electrical-world" boxes (Current, Charge) use a solid double outline, the two "chemical-world" boxes (Moles of electrons, Moles of substance) use a dashed outline, and the final Mass box uses a thick solid outline. Reading order:

Box ① "Current I (amps)" — top-left, solid double outline.

An arrow labelled "× time" points right to Box ② "Charge Q = I × t (C)" — top-middle, solid double outline.

An arrow labelled "÷ F" points right to Box ③ "Moles of electrons = Q / F" — top-right, dashed outline. A side-note reads "F = 96485 C/mol".

A downward arrow labelled "÷ z electrons" drops from Box ③ to Box ④ "Moles of substance = n_e / z" — dashed outline, bottom-right. A side-note reads "z = electrons per atom (Zn, Pb = 2; Li = 1)".

A leftward arrow labelled "× M" points from Box ④ to the final box "Mass = mol × M (grams)" — thick solid outline, bottom-middle. The title across the top reads "The battery tool-chain: C → F → S → M". So the full path is Current →(×time)→ Charge →(÷F)→ Moles of electrons →(÷z)→ Moles of substance →(×M)→ Mass.

Look at the blueprint above — it is the skeleton of every example on this page, so spend a moment on it now and you will never be lost mid-problem. The badges ①②③④ and the outline styles (double-solid = electrical, dashed = chemical, thick-solid = final answer) mean you can follow the sequence by shape and number alone, without depending on colour.

Box ① → Box ② (arrow "× time"): you almost always start with a measured current (amps) and multiply by time to land on charge $Q$ (coulombs). This is tool 1.
Box ② → Box ③ (arrow "÷ F"): dividing charge by Faraday's constant gives moles of electrons. This is tool 2 — the bridge from the electrical world into the chemical world.
Box ③ → Box ④ (downward arrow "÷ z electrons"): here $z$ is how many electrons each atom carries (read it off the half-reaction: $z = 2$ for Zn and Pb, $z = 1$ for Li). Dividing gives moles of substance. This is tool 3.
Box ④ → Mass box (arrow "× M"): multiply moles by the molar mass $M$ to reach mass in grams — usually the answer. This is tool 4.

Notice you can also travel backward along any arrow: if a problem gives you a mass and asks for a time (Ex 5), you enter at the Mass box and walk against the arrows.

Cell A — Basic charge → mass

Step 1 — Total charge. Convert time to seconds, then use tool 1. $t = 2.0 \times 3600 = 7200 s, Q = I \cdot t = 0.30 \times 7200 = 2160 C$ Why this step? Current is charge per second; multiplying by seconds gives the total charge that flowed.

Step 2 — Moles of electrons. Tool 2 (Faraday's bridge). $n_{e^{-}} = \frac{Q}{F} = \frac{2160}{96485} = 0.02239 mol$ Why this step? Each mole of electrons carries $F$ coulombs, so dividing counts how many mole-packets flowed.

Step 3 — Moles of Zn. Tool 3 (stoichiometry): each Zn atom sheds 2 electrons. $n_{Zn} = \frac{n _{e^{-}}}{2} = \frac{0.02239}{2} = 0.011195 mol$ Why divide by 2? The half-reaction shows one Zn produces two electrons, so it takes half as many Zn atoms as electrons.

Step 4 — Mass. Tool 4 (the "M" step). $m = n_{Zn} \times 65.4 = 0.011195 \times 65.4 = 0.732 g$ Why this step? Molar mass converts a count of moles into a weighable mass — the final box on the map.

Verify: Units chain: $A \cdot s = C$ ; $C / (C/mol) = mol$ ; $mol \times g/mol = g$ ✓. The answer $0.73 g$ sits between $0.1$ and $1 g$ — modest, as expected for a small current over 2 h.

Cell B — EMF and the sign trap

Step 1 — Write the master formula (from 2.7.05-Standard-electrode-potentials): $E_{cell} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ Why this step? Both potentials are quoted as reduction potentials. The cathode reduces (used as-is); the anode oxidizes, so we subtract its reduction potential.

Step 2 — Substitute, minding the double negative. $E_{cell} = (+ 1.20) - (- 0.76) = 1.20 + 0.76 = 1.96 V$ Why this step? Subtracting a negative number adds its magnitude — the classic sign trap. If you carelessly wrote $1.20 - 0.76 = 0.44 V$ you would be wrong by the whole zinc contribution.

Verify: The parent note quotes "~2.0 V theoretically", matching $1.96 V$ . Real cells drop to $1.5 V$ from overpotential — consistent. ✓

Cell C — Both electrodes consumed at once

Step 1 — Moles of electrons (tool 2). $n_{e^{-}} = \frac{600}{96485} = 0.006218 mol$ Why this step? The same electrons pass through anode and cathode in series, so both electrodes share this one number.

Step 2 — Anode: moles and mass of Pb. Each Pb gives 2 electrons. $n_{Pb} = \frac{0.006218}{2} = 0.003109 mol, m_{Pb} = 0.003109 \times 207.2 = 0.644 g$ Why this step? The anode half-reaction shows 2 electrons per Pb (tool 3), so we divide the electron moles by 2 to count Pb atoms, then apply tool 4 (× molar mass) to get grams.

Step 3 — Cathode: moles and mass of PbO₂. Each PbO₂ takes 2 electrons. $n_{PbO_{2}} = \frac{0.006218}{2} = 0.003109 mol, m_{PbO_{2}} = 0.003109 \times 239.2 = 0.744 g$ Why this step? Same reasoning as the anode: the cathode half-reaction shows 2 electrons per PbO₂, so divide by 2 (tool 3), then × molar mass (tool 4). We repeat the chain because a lead-acid cell consumes material at both plates simultaneously.

Why different masses? Equal moles ( $0.00311$ ) but $PbO_{2}$ is heavier per mole ( $239.2$ vs $207.2$ ), so its mass is larger. Electrons are shared; grams are not.

Verify: Both use the same $n_{e^{-}}$ and same "÷2", so both give $0.00311 mol$ ✓. The cathode value $0.744 g$ matches the parent note's car-starter example. ✓

Cell D — Charging: run the movie backward

Step 1 — Recognise reversibility. A secondary cell's charging reactions are the discharge reactions run backward with the same electron count. Why this step? Reversing an equation does not change how many electrons ride per formula unit — still 2 per PbO₂.

Step 2 — Reuse Ex 3's chain unchanged. $n_{e^{-}} = \frac{600}{96485} = 0.006218 mol, n_{PbO_{2}} = \frac{0.006218}{2} = 0.003109 mol$ $m_{PbO_{2}} = 0.003109 \times 239.2 = 0.744 g$ Why this step? Because the electron count per PbO₂ is unchanged (still 2), the entire charge → moles-of-electrons → moles-of-PbO₂ → mass chain (tools 2, 3, 4) is numerically identical to discharge; only the direction of travel is reversed. So we simply recopy the same arithmetic rather than deriving anew.

Verify: Identical to Ex 3, so $0.744 g$ ✓. This is the whole point of a rechargeable battery — the same charge in reverse rebuilds exactly what discharge removed (idealised, ignoring inefficiency). Links to reversible thermodynamics in 3.1.12-Gibs-free-energy-and-spontaneity.

Cell E — Lifetime: when a reactant hits zero

This problem gives a mass and asks for a time, so we walk the map backward — enter at the Mass box, travel against every arrow.

Step 1 — Moles of Zn available (tool 4, backward). $n_{Zn} = \frac{4.0}{65.4} = 0.06116 mol$ Why this step? Zn is the limiting reactant; dividing its mass by molar mass (reversing tool 4) converts grams into a count of moles, our entry point on the map.

Step 2 — Total electrons those atoms can supply (tool 3, backward). Two per atom. $n_{e^{-}} = 2 \times 0.06116 = 0.12232 mol$ Why this step? Now we go against the "÷ z" arrow, so we multiply by 2 instead of dividing: each Zn atom can release two electrons, so the total electron budget is twice the Zn count.

Step 3 — Total charge (tool 2, backward). $Q = n_{e^{-}} \times F = 0.12232 \times 96485 = 11802 C$ Why this step? Going backward across the "÷ F" arrow means multiplying by Faraday's constant: each mole of electrons carries $F$ coulombs, so the total charge the zinc can push is electron-moles times $F$ .

Step 4 — Time (tool 1, backward). $t = \frac{Q}{I} = \frac{11802}{0.30} = 39340 s = \frac{39340}{3600} = 10.9 h$ Why this step? Since $Q = I \cdot t$ , solving for time divides total charge by the (constant) current. Dividing seconds by 3600 converts to hours for a human-readable answer.

Verify: Parent note states "~11 hours theoretically, ~8 h in practice." Our $10.9 h$ matches the theoretical figure ✓. The zero-reactant case (battery "dead") is exactly the moment Zn runs out here.

Cell F — Electrolyte dilutes, and EMF drifts

Step 1 — Derive the electron count from the two half-reactions. Write them out and add: $anode, loses 2 e^{-} Pb + SO_{4}^{2 -} \to PbSO_{4} + 2 e^{-} + cathode, gains 2 e^{-} PbO_{2} + 4 H^{+} + SO_{4}^{2 -} + 2 e^{-} \to PbSO_{4} + 2 H_{2} O$ The anode releases exactly $2 e^{-}$ and the cathode absorbs exactly $2 e^{-}$ , so when we add the halves the electrons cancel and the overall reaction transfers $z = 2$ electrons. Why this step? To use tool 2 and the Nernst equation we must know $z$ , and the only honest way to get it is to balance the electrons of the two halves against each other — here both halves already carry $2 e^{-}$ , so no extra scaling is needed and $z = 2$ .

Step 2 — Moles of electrons (tool 2). $n_{e^{-}} = \frac{600}{96485} = 0.006218 mol$ Why this step? Convert the given charge into moles of electrons, the currency the overall equation is written in.

Step 3 — Acid consumed (tool 3, using the overall coefficients). The overall equation shows $2 H_{2} SO_{4}$ consumed per $2 e^{-}$ , a 1:1 ratio of acid-moles to electron-moles. $n_{H_{2} SO_{4}} = \frac{2}{2} \times n_{e^{-}} = 0.006218 mol$ Why this step? We read the coefficients off the balanced overall equation: $2$ acid molecules disappear for every $2$ electrons that flow, giving the 1:1 factor we multiply by.

Step 4 — Density direction. Concentrated $H_{2} SO_{4}$ is dense ( $\approx 1.25 g/cm^{3}$ ); replacing it with water ( $1.00 g/cm^{3}$ ) lowers density. Why this step? Density is what a hydrometer reads. Since discharge removes heavy acid and adds light water, the mixture's density must fall.

Step 5 — EMF drift via the Nernst equation (the Nernst-flavoured heart of this case). From 2.7.06-Nernst-equation-and-concentration-effects, for the discharge cell $E = E^{\circ} - \frac{0.0592}{z} lo g (\frac{1}{a _{H_{2} SO_{4}}^{2}}) = E^{\circ} + \frac{0.0592}{z} \cdot 2 lo g a_{H_{2} SO_{4}}$ with $z = 2$ (from Step 1) and $E^{\circ} = 2.05 V$ . The reactant $H_{2} SO_{4}$ has coefficient $2$ , so its activity appears squared. As the acid is consumed, $a_{H_{2} SO_{4}}$ falls, so $lo g a_{H_{2} SO_{4}}$ becomes more negative and $E$ drops. If the activity falls to half its start value, $a \to 0.5$ : $Δ E = \frac{0.0592}{2} \cdot 2 lo g (0.5) = 0.0592 \times (- 0.301) = - 0.0178 V$ Why this tool and not another? The Nernst equation is the only tool that converts a concentration change into a voltage change — exactly the question "how much does the EMF drift as the acid dilutes?". A plain stoichiometry step tells us how much acid is gone but says nothing about voltage; Nernst closes that gap. The drop of about $18 mV$ (per halving) is why a tired car battery reads slightly below its fresh voltage even before it is fully flat.

Verify: Parent note: charged $ρ \approx 1.25$ , discharged $ρ \approx 1.15 g/cm^{3}$ — density drops on discharge ✓. Acid consumed $n_{H_{2} SO_{4}} = n_{e^{-}} = 0.006218 mol$ ✓. Nernst drift for $a = 0.5$ : $\frac{0.0592}{2} \cdot 2 lo g (0.5) = - 0.0178 V$ ✓ (negative → EMF sags, matching physical intuition).

Cell G — Real-world word problem

Step 1 — Charge in one second. $Q = I \cdot t = 2.0 \times 1.0 = 2.0 C$ Why this step? "Per second" means set $t = 1 s$ ; then $Q = I \cdot t$ gives the charge delivered in that single second.

Step 2 — Count electrons directly using $e$ (not $F$ , since we want individual electrons). $N = \frac{Q}{e} = \frac{2.0}{1.602 \times 1 0 ^{- 19}} = 1.25 \times 1 0^{19} electrons$ Why this step? Here we want a raw count of electrons, not moles, so we divide by the charge of a single electron $e$ rather than by $F$ (which would give moles). Each electron carries $e$ coulombs, so total charge divided by per-electron charge counts them.

Verify: Order of magnitude $1 0^{19}$ ✓ — larger than the flashlight's $1.9 \times 1 0^{18}$ because current is bigger. Cross-check via Faraday: $2.0/96485 = 2.07 \times 1 0^{- 5} mol$ , times $N_{A}$ gives $1.25 \times 1 0^{19}$ ✓ (both routes agree).

Cell H — Exam twist: 1-electron ion & mAh

Step 1 — Convert mAh to coulombs. $1 mAh = 1 0^{- 3} A \times 3600 s = 3.6 C$ . $Q = 2000 \times 3.6 = 7200 C$ Why this step? mAh (milliamp-hours) is a charge unit in disguise: current × time. We convert to coulombs so our standard tools apply.

Step 2 — Moles of electrons (tool 2). $n_{e^{-}} = \frac{7200}{96485} = 0.07463 mol$ Why this step? Divide the charge by $F$ to cross from the electrical world into moles of electrons.

Step 3 — Moles of Li (tool 3). One electron per Li, so ÷1. $n_{Li} = \frac{0.07463}{1} = 0.07463 mol$ Why this step? The half-reaction $Li^{+} + e^{-} \to Li$ has $z = 1$ , so every electron moves exactly one Li atom — dividing by 1 leaves the count unchanged. Same charge therefore moves twice as many Li atoms as a divalent metal would.

Step 4 — Mass (tool 4). Molar mass $M_{Li} = 6.94 g/mol$ . $m = n_{Li} \times M_{Li} = 0.07463 \times 6.94 = 0.518 g$ Why this step? The final "× M" arrow on the map: multiplying moles by molar mass converts the count of Li atoms into a weighable mass in grams.

Verify: Units: $mAh \to C \to mol \to g$ ✓. Sanity: lithium is very light ( $6.94 g/mol$ ) and single-electron, which is exactly why Li-ion packs so much charge per gram — half a gram stores $2000 mAh$ . ✓

Cell I — Degenerate edge cases

Part (a) — Zero current. $Q = I \cdot t = 0 \times 86400 = 0 C \Rightarrow m_{Zn} = 0 g$ Why this step? With no closed circuit the current is exactly zero, so tool 1 multiplies by zero at the very first box of the map — every downstream quantity is therefore zero too. (In reality tiny self-discharge occurs, but the electrochemical consumption from delivered current is exactly zero.)

Part (b) — Fully depleted. Ex 5 found total deliverable charge $= 11802 C$ . Once all zinc is gone, remaining deliverable charge is $Q_{left} = 11802 - 11802 = 0 C$ Why this step? The zinc budget from Ex 5 caps how much charge the cell can ever deliver; subtracting the charge already delivered from that ceiling gives what is left. When it reaches zero the reactant is fully spent.

Why this matters: This is the limiting boundary — the cell EMF collapses because a reactant concentration has hit zero, driving the Nernst term to negative infinity (2.7.06-Nernst-equation-and-concentration-effects). The battery is "dead."

Verify: Both degenerate answers are $0$ ✓. These bracket the physically meaningful range: $0 \leq Q_{delivered} \leq 11802 C$ .

Seeing it all at once

The bar chart below puts the numeric answers from several examples side by side, so you can feel the relative scales — why half a gram of lithium (Ex 8) stores so much more charge-per-gram than the grams of lead an equal charge moves (Ex 3).

Alt text / full description: A blueprint-blue bar chart with a light white horizontal grid. The vertical axis is "mass consumed / regenerated (g)" running 0 to about 0.9. Four bars, each outlined (not filled) and topped with its numeric value: Zn (Ex 1, 2160 C) = 0.732 g, Pb (Ex 3, 600 C) = 0.644 g, PbO₂ (Ex 3, 600 C) = 0.744 g, and Li (Ex 8, 7200 C) = 0.518 g. The three metal-plate bars (Zn, Pb, PbO₂) share one outline colour; the lithium bar is drawn in the accent colour and is the shortest, driving home its mass efficiency.

Notice the amber lithium bar: for a modest half-gram it corresponds to a full $2000 mAh$ — lithium's tiny molar mass and single-electron transfer make it the mass-efficiency champion, the physical reason Li-ion batteries dominate portable electronics.

Recall

Recall The universal chain — fill the blanks

Current × time gives ::: charge $Q = I t$ (tool 1) Charge ÷ Faraday's constant gives ::: moles of electrons (tool 2) Moles of electrons ÷ (electrons per atom $z$ ) gives ::: moles of substance (tool 3) Moles × molar mass gives ::: mass in grams (tool 4, the "M" step) $E_{cell} = E_{cathode} - E_{anode}$ and subtracting a negative potential means you ::: add its magnitude Zn and Pb transfer how many electrons each? ::: 2 Li⁺ → Li transfers how many electrons? ::: 1 During lead-acid discharge the acid density and EMF both ::: fall (acid consumed → water made → Nernst term drops)

See also: 2.7.09-Corosion-and-prevention for the unwanted redox cousin of these batteries.