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

Intuition How to use this page

Cover the answer, try the problem, then open the solution. Every problem below is a small piece of the machinery you built in the parent note. We climb five rungs: first just recognising the parts, then using one formula, then analysing what changes, then combining ideas, and finally mastering a multi-step design question.

Two tools appear over and over, so let us pin them down in plain words before any symbol shows up:

Charge is "how much electricity flowed". Its unit is the coulomb (C). A wire carrying a steady current (charge per second, measured in amperes, $A = C/s$ ) for some seconds delivers charge $Q = I \cdot t$ .
The Faraday constant $F = 96485 C/mol$ is the charge carried by one mole of electrons. It is the bridge between the electrical world (coulombs) and the chemical world (moles). Whenever a problem hands you amps and seconds but asks for grams, $F$ is the door between the two rooms.

The figure below is your master map for every numerical problem on this page. Look at it now: the left (orange) box is the "electrical room" where a problem gives you amps and seconds, which you multiply into charge $Q$ . The arrow through the middle is the Faraday door — dividing by $F$ carries you into the right (violet) "chemical room" of moles. The small violet note at the bottom is the one thing students forget: the electron count $z$ , read straight off the half-reaction, links moles-of-electrons to moles-of-substance. Nearly every solution below is just a walk left-to-right across this picture.

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

Prerequisites you may want open in another tab: 2.7.01-Oxidation-numbers-and-redox-definition, 2.7.04-Galvanic-cells-and-cell-notation, 2.7.05-Standard-electrode-potentials, 2.7.06-Nernst-equation-and-concentration-effects, 3.1.12-Gibs-free-energy-and-spontaneity.

Level 1 — Recognition

Exercise 1.1

Classify each as primary or secondary, and say in one word why: (a) a car lead-acid battery, (b) a torch/flashlight dry cell, (c) a phone Li-ion cell.

Recall Solution

(a) Secondary — rechargeable. Its redox reactions reverse cleanly when you push external voltage in.
(b) Primary — single-use. The Leclanché chemistry is effectively irreversible once the Zn can is eaten.
(c) Secondary — rechargeable. Li⁺ ions shuttle back and forth (intercalation) many hundreds of cycles.

Exercise 1.2

In the dry cell, name the material at (a) the anode, (b) the cathode, and state which one is oxidised.

Recall Solution

(a) Anode = zinc metal (Zn), the outer can. This is where oxidation happens: $Zn \to Zn^{2 +} + 2 e^{-}$ .
(b) Cathode = MnO₂ (packed around a graphite current-collector rod). Here reduction happens.
The anode is always the oxidation electrode, in every cell, by definition. Remember the vowel trick below.

Level 2 — Application

Exercise 2.1

A dry cell drives a current of $0.20 A$ for $30 minutes$ . How many coulombs of charge pass, and how many electrons is that? (Use $e = 1.6 \times 1 0^{- 19} C$ .)

Recall Solution

Step 1 — get the time into seconds. Why: current is coulombs per second, so time must be in seconds. $t = 30 min \times 60 = 1800 s$ Step 2 — charge from $Q = I t$ . Why: this is the definition of current rearranged. $Q = 0.20 A \times 1800 s = 360 C$ Step 3 — electrons = total charge ÷ charge per electron. Why: each electron carries $1.6 \times 1 0^{- 19} C$ , so the number of them is the total coulombs divided by the coulombs-per-electron. $N = \frac{360}{1.6 \times 1 0 ^{- 19}} = 2.25 \times 1 0^{21} electrons$

Exercise 2.2

How many grams of zinc are consumed by the current in Exercise 2.1? ( $M_{Zn} = 65.4 g/mol$ , $F = 96485 C/mol$ , and $Zn \to Zn^{2 +} + 2 e^{-}$ .)

Recall Solution

Step 1 — moles of electrons via the Faraday door. Why: $F$ converts coulombs into moles-of-electrons. $n_{e^{-}} = \frac{Q}{F} = \frac{360}{96485} = 3.731 \times 1 0^{- 3} mol$ Step 2 — moles of Zn. Why: each Zn atom releases 2 electrons, so it takes 2 mol $e^{-}$ to dissolve 1 mol Zn — divide by 2. $n_{Zn} = \frac{3.731 \times 1 0 ^{- 3}}{2} = 1.866 \times 1 0^{- 3} mol$ Step 3 — mass. Why: molar mass $M$ is by definition the grams contained in one mole, so multiplying moles by $M$ converts a count of moles straight into a mass in grams. $m = n_{Zn} \times M_{Zn} = 1.866 \times 1 0^{- 3} \times 65.4 = 0.122 g$

Level 3 — Analysis

Exercise 3.1

Recall the lead-acid discharge reaction: $Pb + PbO_{2} + 2 H_{2} SO_{4} \to 2 PbSO_{4} + 2 H_{2} O$ As the battery discharges, does the electrolyte get denser or less dense? Explain using the equation, and say how a mechanic could measure the state of charge.

Recall Solution

What the equation says: two molecules of dense H₂SO₄ are consumed and two molecules of water are produced per cycle of the reaction. Consequence: the acid concentration drops and water content rises, so the electrolyte becomes less dense as it discharges. (Charged $ρ \approx 1.25 g/cm^{3}$ ; discharged $ρ \approx 1.15 g/cm^{3}$ .) Measurement: a hydrometer — a weighted float — sits higher in denser liquid. Reading how high it floats gives the acid density directly, hence the state of charge. This is why "the battery is flat" often literally means "the acid has thinned out".

Exercise 3.2

The parent note gives the lead-acid half-cells with the cathode as a reduction potential $E_{cathode}^{\circ} = + 1.69 V$ , but writes the anode as an oxidation potential $E_{ox}^{\circ} = + 0.36 V$ for $Pb + SO_{4}^{2 -} \to PbSO_{4} + 2 e^{-}$ . Show that the standard cell EMF comes out to $2.05 V$ , and explain why we must not just add $1.69 + 0.36$ blindly without first checking which value is a reduction and which is an oxidation.

Recall Solution

The clean rule: $E_{cell}^{\circ} = E_{cathode(red)}^{\circ} - E_{anode(red)}^{\circ}$ , using both as reduction potentials. The anode half-reaction is written as an oxidation with $E_{ox}^{\circ} = + 0.36 V$ ; its reduction potential is the same number with the sign flipped, i.e. $E_{anode(red)}^{\circ} = - 0.36 V$ . $E_{cell}^{\circ} = 1.69 - (- 0.36) = 2.05 V$ Why not just add? Adding $1.69 + 0.36$ happens to give the same number here only because the anode value was already flipped to the oxidation convention. If you mix conventions carelessly you can accidentally subtract twice. Safe habit: convert everything to reduction potentials, then do cathode minus anode. Six such cells in series give $6 \times 2.05 \approx 12.3 V$ — the familiar "12-volt" car battery.

Level 4 — Synthesis

Exercise 4.1

A car starter motor draws $150 A$ for $4.0 s$ . During this burst: (a) find the charge, (b) the moles of electrons, (c) the mass of Pb oxidised at the anode, and (d) the mass of H₂SO₄ consumed overall. ( $M_{Pb} = 207.2$ , $M_{H_{2} SO_{4}} = 98.1 g/mol$ .)

Recall Solution

Step 1 — charge. $Q = I t = 150 \times 4.0 = 600 C$ . Step 2 — moles of electrons. $n_{e^{-}} = \frac{600}{96485} = 6.219 \times 1 0^{- 3} mol$ . Step 3 — mass of Pb. The anode half-reaction $Pb + SO_{4}^{2 -} \to PbSO_{4} + 2 e^{-}$ has $z = 2$ . $n_{Pb} = \frac{6.219 \times 1 0 ^{- 3}}{2} = 3.109 \times 1 0^{- 3} mol \Rightarrow m_{Pb} = 3.109 \times 1 0^{- 3} \times 207.2 = 0.644 g$ Step 4 — H₂SO₄ consumed. Why this ratio: the overall equation $Pb + PbO_{2} + 2 H_{2} SO_{4} \to 2 PbSO_{4} + 2 H_{2} O$ shows 2 mol H₂SO₄ used for each full turn of the reaction, and one full turn transfers 2 mol electrons (the anode and cathode each pass $2 e^{-}$ ). So the ratio is $2 H_{2} SO_{4} : 2 e^{-} = 1 : 1$ — moles of H₂SO₄ equal moles of electrons. $n_{H_{2} SO_{4}} = 6.219 \times 1 0^{- 3} mol \Rightarrow m = 6.219 \times 1 0^{- 3} \times 98.1 = 0.610 g$ Reading it back: a whole cell holds ~1 kg of active material, so this $0.6 g$ burst is trivially sustainable — exactly why the same battery starts your car thousands of times.

Exercise 4.2

Connect chemistry to spontaneity: for one lead-acid cell $E_{cell}^{\circ} = 2.05 V$ with $z = 2$ electrons transferred. Compute the standard Gibbs free energy change $Δ G^{\circ} = - z F E_{cell}^{\circ}$ in kJ per mole of reaction, and say what its sign tells you.

Recall Solution

Why this formula: the maximum electrical work a cell can do equals the free energy it releases; $Δ G^{\circ} = - z F E^{\circ}$ is that statement (see 3.1.12-Gibs-free-energy-and-spontaneity). $Δ G^{\circ} = - (2) (96485) (2.05) = - 3.956 \times 1 0^{5} J/mol = - 395.6 kJ/mol$ Meaning of the sign: $Δ G^{\circ} < 0$ ⇒ the discharge reaction is spontaneous — the cell wants to run and push electrons out. To recharge you must supply more energy than this to reverse it, which is why charging always needs an external voltage above $2.05 V$ .

Level 5 — Mastery

Exercise 5.1

Design/reasoning question. A phone Li-ion pack is rated $3000 mAh$ at $3.7 V$ . (a) Convert its capacity to coulombs. (b) How many moles of electrons can it deliver? (c) If the cathode uses the couple with $z = 1$ electron per Li⁺ shuttled, how many moles of Li⁺ move across per full discharge? (d) Estimate the total electrical energy stored in joules and watt-hours, and (e) explain why watt-hours, not amp-hours, is the honest way to compare batteries of different voltages.

Recall Solution

(a) Capacity → coulombs. $3000 mAh = 3.0 Ah$ . Amp-hours means amps × hours, and $1 h = 3600 s$ : $Q = 3.0 A \times 3600 s = 10800 C$ (b) Moles of electrons. Faraday door again: $n_{e^{-}} = \frac{10800}{96485} = 0.1119 mol$ (c) Moles of Li⁺. With $z = 1$ , each electron corresponds to one Li⁺ shuttled, so $n_{Li^{+}} = 0.1119 mol .$ (d) Energy. Energy = charge × voltage (this is what a volt means: joules per coulomb): $E = Q V = 10800 \times 3.7 = 39960 J \approx 4.00 \times 1 0^{4} J$ In watt-hours ( $1 Wh = 3600 J$ ): $\frac{39960}{3600} = 11.1 Wh$ . (Equivalently $3.0 Ah \times 3.7 V = 11.1 Wh$ .) (e) Why watt-hours are honest: amp-hours count charge only, ignoring the "height of the waterfall" (voltage). A $3 Ah$ cell at $3.7 V$ stores far more usable energy than a $3 Ah$ cell at $1.5 V$ . Watt-hours multiply charge by voltage, so they compare the actual energy — the fair currency. This is the numerical face of the "waterfall split across two cliffs" picture from the parent note.

Recall Rapid self-check (cloze)

Charge from current and time ::: $Q = I t$ The Faraday constant equals ::: $96485 C per mole of electrons$ Faraday's mass law ::: $m = \frac{Q M}{z F}$ Free energy from cell EMF ::: $Δ G^{\circ} = - z F E_{cell}^{\circ}$ Energy stored from charge and voltage ::: $E = Q V$ Fair way to compare batteries of different voltage ::: watt-hours, not amp-hours