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

Core Concepts

[!definition] Primary vs Secondary Batteries

• ==Primary batteries== are single-use (non-rechargeable). Once the chemical reactants are consumed, the redox reaction stops and the battery is dead. The reactions are essentially irreversible under normal conditions.
• ==Secondary batteries== are rechargeable. The redox reactions are reversible—applying external voltage runs the reaction backward, regenerating the reactants. You can cycle them hundreds to thousands of times.

Why this distinction matters: Primary batteries have simpler chemistry (cheaper, more stable shelf life) but create waste. Secondary batteries require more sophisticated electrode materials that maintain structure through repeated redox cycling.

1. Primary Battery: The Dry Cell (Leclanché Cell)

Structure & Components

A dry cell uses a paste electrolyte (not liquid), making it leak-resistant and portable.

Components:

• Anode (−): Zinc metal can (outer casing) — gets oxidized
• Cathode (+): Graphite rod surrounded by MnO₂ (manganese dioxide) mixed with carbon powder
• Electrolyte: Paste of NH₄Cl (ammonium chloride) and ZnCl₂, sometimes with starch as thickener
• Separator: Porous paper soaked in electrolyte, prevents direct contact between electrodes

The Chemistry — Derived from First Principles

At the anode (oxidation): Zinc metal loses electrons: $\text{Zn}(s) \to \text{Zn}^{2+}(aq) + 2e^- \quad E^\circ_{\text{ox}} = +0.76\,\text{V}$

Why does Zn want to oxidize? Zinc sits high on the activity series—it loses electrons easily to form stable Zn²⁺ ions. The standard reduction potential of Zn²⁺/Zn is −0.76 V, so oxidation gives +0.76 V.

At the cathode (reduction): This is where it gets interesting. The cathode reaction is complex and involves multiple steps:

1. Primary reduction: MnO₂ is reduced by electrons from the external circuit: $2\text{MnO}_2(s) + 2\text{H}^+(aq) + 2e^- \to \text{Mn}_2\text{O}_3(s) + \text{H}_2\text{O}(l)$

2. Why H⁺? The amonium chloride electrolyte provides protons through hydrolysis: $\text{NH}_4^+(aq) \rightleftharpoons \text{NH}_3(aq) + \text{H}^+(aq)$

3. Secondary reaction: Zinc ions from the anode react with ammonia and chloride: $\text{Zn}^{2+}(aq) + 2\text{NH}_3(aq) + 2\text{Cl}^-(aq) \to [\text{Zn}(\text{NH}_3)_2]\text{Cl}_2(s)$

This forms a complex salt that precipitates, preventing Zn²⁺ buildup (which would slow the reaction via Le Chatelier's principle).

Overall approximate reaction: $\text{Zn}(s) + 2\text{MnO}_2(s) + 2\text{NH}_4^+(aq) \to \text{Zn}^{2+}(aq) + \text{Mn}_2\text{O}_3(s) + 2\text{NH}_3(aq) + \text{H}_2\text{O}(l)$

Cell voltage: ~1.5 V initially, drops to ~0.9 V as reactants deplete.

[!example] Worked Example: Electron Flow in a Flashlight

Setup: A flashlight uses two1.5 V dry cells in series (3.0 V total) to light a 0.3 A bulb.

Step 1 — Calculate electrons flowing per second: Current I = 0.3 A = 0.3 coulombs/second. One electron carries $e = 1.6 \times 10^{-19}$ C. $\text{Number of electrons/s} = \frac{0.3}{1.6 \times 10^{-19}} = 1.875 \times 10^{18} \,e^-/\text{s}$

Why this step? Current is defined as charge per time. We convert to fundamental charge carriers.

Step 2 — How much Zn is consumed per hour? Each Zn atom releases2 electrons: $\text{Zn} \to \text{Zn}^{2+} + 2e^-$ $\text{Zn atoms consumed/s} = \frac{1.875 \times 10^{18}}{2} = 9.375 \times 10^{17} \,\text{atoms/s}$

Moles per second: $\frac{9.375 \times 10^{17}}{6.022 \times 10^{23}} = 1.557 \times 10^{-6}\,\text{mol/s}$

Per hour: $1.557 \times 10^{-6} \times 3600 = 5.6 \times 10^{-3}\,\text{mol} = 0.0056\,\text{mol}$

Mass of Zn (M = 65.4 g/mol): $0.0056 \times 65.4 = 0.366\,\text{g}$

Why this matters: A typical AA dry cell has ~4 g of Zn, so it could theoretically run flashlight for ~11 hours, though voltage drop and incomplete reaction reduce this to ~8 hours in practice.

2. Secondary Battery: Lead-Acid

The lead-acid battery (invented 1859, still in cars today!) uses reversible redox between lead compounds.

Structure

• Anode (−): Spongy lead (Pb) plates
• Cathode (+): Lead dioxide (PbO₂) plates
• Electrolyte: Concentrated sulfuric acid (H₂SO₄), ~4.5 M

Chemistry — Discharge (providing power)

At anode (oxidation): $\text{Pb}(s) + \text{SO}_4^{2-}(aq) \to \text{PbSO}_4(s) + 2e^- \quad E^\circ = +0.36\,\text{V}$

Why does lead form PbSO₄? Sulfate ions are abundant in the electrolyte. Lead(II) sulfate is sparingly soluble—it precipitates onto the electrode surface, maintaining electrical contact.

At cathode (reduction): $\text{PbO}_2(s) + 4\text{H}^+(aq) + \text{SO}_4^{2-}(aq) + 2e^- \to \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \quad E^\circ = +1.69\,\text{V}$

Why this direction? PbO₂ is lead in +4 oxidation state—it's a strong oxidizer (wants electrons). In acidic solution, it grabs electrons and protons, reducing to Pb²⁺ which precipitates as PbSO₄.

Overall discharge reaction: $\text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \to 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$

Key observations during discharge:

1. Both electrodes form PbSO₄ (white coating)
2. Sulfuric acid is consumed — concentration decreases
3. Water is produced — electrolyte becomes more dilute
4. EMF: $E_{\text{cell}} = 1.69 - 0.36 = 2.0\,\text{V}$ per cell (car batteries have6 cells = 12 V)

Chemistry — Charging (reversing the reaction)

Apply external voltage >2.0 V per cell. Every reaction runs backward:

At cathode during charging (now oxidation): $\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \to \text{PbO}_2(s) + 4\text{H}^+(aq) + \text{SO}_4^{2-}(aq) + 2e^-$

At anode during charging (now reduction): $\text{PbSO}_4(s) + 2e^- \to \text{Pb}(s) + \text{SO}_4^{2-}(aq)$

Result:

• PbSO₄ converts back to Pb and PbO₂
• H₂SO₄ concentration increases
• Water is consumed

[!example] Worked Example: Starting a Car

Problem: A car starter motor draws 200 A for 3 seconds. How much PbO₂ is consumed at the cathode?

Step 1 — Calculate total charge: $Q = I \cdot t = 200\,\text{A} \times 3\,\text{s} = 600\,\text{C}$

Why this step? Charge = current × time by definition. One mole of electrons = 96,485 C (Faraday's constant).

Step 2 — Moles of electrons: $n_{e^-} = \frac{600}{96,485} = 0.00622\,\text{mol}$

Step 3 — Stoichiometry of cathode reaction: $\text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \to \text{PbSO}_4 + 2\text{H}_2\text{O}$ Each PbO₂ consumes 2 electrons, so: $n_{\text{PbO}_2} = \frac{0.00622}{2} = 0.00311\,\text{mol}$

Why divide by 2? The balanced equation shows 2 e⁻ per PbO₂. Stoichiometry must be respected.

Step 4 — Mass of PbO₂: Molar mass of PbO₂ = 207.2 + 2(16) = 239.2 g/mol $m = 0.00311 \times 239.2 = 0.744\,\text{g}$

Interpretation: A car battery has ~1kg of PbO₂ per cell, so this 3-second draw consumes <0.1% of one cell's material—easily sustainable.

3. Secondary Battery: Lithium-Ion

The lithium-ion battery (1991, revolutionized portable electronics) uses intercalation chemistry—Li⁺ ions shuttle between layered materials without destroying the crystal structure.

Structure (LiCoO₂ cathode variant)

• Anode (−): Graphite layers that can host Li atoms between planes
• Cathode (+): Lithium cobalt oxide (LiCoO₂) — layered oxide
• Electrolyte: Organic solvent (e.g., ethylene carbonate) with LiPF₆ salt — conducts Li⁺, not electrons
• Separator: Porous polymer film (prevents short circuit but passes ions)

Chemistry — Discharge

At anode (oxidation): Lithium atoms deintercalate (leave) the graphite: $\text{Li}_x\text{C}_6 \to x\text{Li}^+ + xe^- + \text{C}_6$ Typically $x \approx 1$, giving $\text{LiC}_6$ when fully charged.

Why graphite? Its layered structure has van der Waals gaps that accommodate Li atoms reversibly. The carbon planes conduct electrons while Li⁺ sits between them.

At cathode (reduction): Li⁺ ions intercalate (insert) into the LiCoO₂ lattice: $\text{Li}_{1-x}\text{CoO}_2 + x\text{Li}^+ + xe^- \to \text{LiCoO}_2$

Why CoO₂? Cobalt cycles between Co³⁺ and Co⁴⁺ oxidation states, accepting electrons when Li⁺ inserts. The layered structure keeps Co-O planes intact during cycling.

Overall discharge: $\text{LiC}_6 + \text{Li}_{1-x}\text{CoO}_2 \to \text{C}_6 + \text{LiCoO}_2$

Voltage: ~3.7 V per cell (much higher energy density than lead-acid's 2V!)

Why high voltage? Large difference in chemical potential between lithium (very electropositive, wants to lose electrons) and cobalt(IV) oxide (strong oxidizer, wants electrons).

Charging

Reverse every step:

• Li⁺ ions leave LiCoO₂ cathode (now oxidation): $\text{LiCoO}_2 \to \text{Li}_{1-x}\text{CoO}_2 + x\text{Li}^+ + xe^-$
• Li⁺ ions insert into graphite anode (now reduction): $\text{C}_6 + x\text{Li}^+ + xe^- \to \text{Li}_x\text{C}_6$

[!example] Worked Example: Phone Battery Capacity

Specs: A phone battery is rated 3000 mAh at 3.7 V.

Step 1 — Calculate stored energy: $E = Q \times V = (3000\,\text{mAh}) \times (3.7\,\text{V})$ Convert mAh to coulombs: $3000\,\text{mAh} = 3\,\text{Ah} = 3 \times 3600,\text{C} = 10{,}800\,\text{C}$ $E = 10{,}800\,\text{C} \times 3.7\,\text{V} = 39{,}960\,\text{J} \approx 11.1\,\text{Wh}$

Why this step? Energy = charge × voltage. We convert units to get watt-hours, the standard for battery energy.

Step 2 — How much lithium moves during one full cycle? Total charge: 10,800 C Moles of electrons: $n = \frac{10{,}800}{96{,}485} = 0.112\,\text{mol}$

Each Li atom provides 1 electron, so 0.112 mol of Li cycles. Mass: $0.112\,\text{mol} \times 6.94\,\text{g/mol} = 0.777\,\text{g}$

Why this matters: Less than 1 gram of lithium cycling stores enough energy to run your phone all day. Compare this to ~50 g of Pb consumed in a car battery for the same energy—that's the power of high voltage + lightweight materials.

Common Misconceptions

Practice Problems

Connections

• 2.7.01-Oxidation-numbers-and-redox-definition — Batteries are practical galvanic cells; identify oxidation states changing
• 2.7.04-Galvanic-cells-and-cell-notation — Battery structure follows galvanic cell design (anode/cathode/electrolyte/separator)
• 2.7.05-Standard-electrode-potentials — EMF calculated from half-cell potentials; predicts feasibility
• 2.7.06-Nernst-equation-and-concentration-effects — Battery voltage drops as reactants deplete (Q increases)
• 2.7.09-Corosion-and-prevention — Zinc casing corrodes (intentional oxidation), analogous to unwanted corosion
• 3.1.12-Gibs-free-energy-and-spontaneity — $\Delta G = -nFE_{\text{cell}}$ connects battery voltage to thermodynamic spontaneity

#flashcards/chemistry

What defines a primary battery? :: A non-rechargeable battery where the redox reaction is essentially irreversible under normal conditions. Once reactants are consumed, the battery is dead (e.g., dry cell).

Why does a dry cell produce 1.5 V? ::: The potential difference between Zn oxidation (+0.76 V) and MnO₂ reduction in acidic medium (~+1.2 V theoretical) gives ~2 V, but real overpotential losses yield 1.5 V in practice.

Write the overall discharge reaction for a lead-acid battery :: $\text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \to 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$. Both electrodes form lead sulfate.

What forms on both electrodes of a discharging lead-acid battery? :: Lead sulfate (PbSO₄), a white precipitate. This is unusual—most batteries form different compounds at each electrode.

Why do lithium-ion batteries have higher energy density than lead-acid? :: Higher voltage per cell (~3.7 V vs. 2.0 V) and lightweight lithium (M = 6.94 g/mol) compared to heavy lead (M = 207.2 g/mol). More energy per unit mass.

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)