CLASS 12 CHEMISTRY · CHAPTER 2 · STUDY NOTES
Electrochemistry
A comprehensive study of the relationship between chemical energy and electrical energy, covering galvanic and electrolytic cells, the Nernst equation, conductance of ionic solutions, and practical applications like batteries and fuel cells.
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Introduction to Electrochemical Cells
Follow the electron circuit and the ion circuit separately.Let a reaction drive a circuit—or drive it backwards
- Electrochemical cell
- A system coupling oxidation and reduction at electrodes with charge transport through electronic and ionic conductors.
| Opposing source | Ideal behaviour | Electron path | Net chemistry |
|---|---|---|---|
| Below 1.10 V | Forward galvanic cell | Zn → Cu through load | Zn dissolves; Cu deposits |
| Equals 1.10 V | Balanced opposing emf | No net external current | No net conversion; microscopic exchange continues |
| Above 1.10 V | Driven reverse reaction | Cu → source → Zn | Cu dissolves; Zn deposits |
A galvanic cell converts the chemical driving force of a spontaneous redox reaction into electrical work. In the Daniell cell Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), zinc is oxidised at the negative anode and copper ions are reduced at the positive cathode. The separated half-reactions let electrons pass through an external load rather than directly between reactants.
An electrolytic cell uses an external source to drive a non-spontaneous reaction. Opposing the ideal Daniell emf with a source below 1.10 V leaves the forward direction, equality gives zero net current, and a larger opposing voltage can reverse the reaction. Then copper is the positive anode and zinc the negative cathode. Oxidation remains anodic and reduction cathodic even when signs change.
Let’s look a little closer.
Go deeper: The two charge paths must join
The wire carries electrons. The electrolyte and salt bridge carry ions, with bridge anions moving into the zinc compartment and cations into the copper compartment during the forward reaction. The bridge maintains near-electroneutrality and completes the circuit; electrons do not travel through the bridge.
The opposing-voltage comparison is ideal. A practical operating voltage also involves internal resistance and electrode overpotential. Zero net reaction rate at balanced voltage does not require molecular motion or opposing microscopic reactions to stop.
Take a moment. What do you think?
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Question
An ideal Daniell cell is opposed by 1.40 V. Identify the anode, cathode and electrode signs if the reaction is driven backwards.
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Galvanic Cells and Electrode Potential
Subtract reduction potentials; do not add arbitrary electrode signs.Two half-cells, one continuous charge circuit
- Cell emf
- The reversible open-circuit potential difference between the electrodes; for the written cell Ecell = Eright − Eleft.
For a spontaneous cell written with oxidation on the left and reduction on the right, E°cell = E°cathode − E°anode, using reduction potentials for both half-cells. Zn²⁺/Zn has −0.76 V and Cu²⁺/Cu +0.34 V, so the standard Daniell emf is 1.10 V. A more positive reduction potential favours reduction relative to a partner.
Cell notation Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s) uses a single bar for a phase boundary and a double bar for the salt bridge. In Cu/Ag, Cu → Cu²⁺+2e⁻ and 2Ag⁺+2e⁻→2Ag give Cu+2Ag⁺→Cu²⁺+2Ag, with standard emf 0.80−0.34 = 0.46 V.
Go deeper: Balance electrons without scaling electrode potential
Multiplying a half-equation scales its reaction amount and Gibbs energy, but not its intensive electrode potential. Higher standard reduction potentials identify stronger oxidants among the oxidised forms under standard conditions; the reduced form of a more negative couple is often the stronger reductant.
Only a potential difference can be measured directly. The potential of a single electrode is assigned relative to a reference electrode. Loaded terminal voltage can differ from reversible emf because current introduces losses.
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Question
A metal X²⁺/X has E° = −0.20 V and Ag⁺/Ag has +0.80 V. Write the balanced cell reaction and calculate its standard emf.
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Measurement of Electrode Potential
The hydrogen electrode is a reference, not a source of absolute voltage.A reference electrode fixes the zero
- Standard hydrogen electrode (SHE)
- A platinum-based hydrogen electrode assigned E° = 0 V, with H₂ at standard pressure and H⁺ at unit activity.
Platinum coated with platinum black contacts aqueous H⁺ and hydrogen gas. For standard conditions, H₂ pressure is 1 bar and H⁺ activity is 1; an ideal concentration treatment approximates this as 1 mol/L H⁺. The reversible half-reaction is 2H⁺ + 2e⁻ ⇌ H₂. Platinum conducts and catalyses the interfacial reaction.
The standard potential is defined as zero at each temperature. When coupled to Cu²⁺/Cu under standard conditions, copper is reduced and SHE is oxidised, producing a +0.34 V cell emf. Coupled to zinc, SHE is reduced while zinc is oxidised; E°cell = 0−(−0.76) = +0.76 V.
Go deeper: Reference conditions matter
The activity describes effective chemical concentration relative to a standard state. Exactly 1 mol/L is not automatically activity 1 in a non-ideal solution. Gas pressure, temperature and composition must be specified before comparing measured values.
A hydrogen electrode away from standard conditions need not be at zero potential. Its Nernst dependence follows the activities and gas pressure; the convention concerns the standard potential, not every hydrogen electrode measurement.
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Question
A standard test electrode is the cathode against SHE and the measured emf is +0.52 V. What is its standard reduction potential? Would a non-standard hydrogen electrode necessarily be zero?
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The Nernst Equation
Build the balanced reaction quotient before calculating voltage.Composition changes the cell’s driving force
- Nernst equation
- E = E° − (RT/nF)ln Q for the balanced cell reaction, using dimensionless activities in Q.
At 298 K the base-10 form is E = E°−(0.05916 V/n)log₁₀Q. For Zn+Cu²⁺→Zn²⁺+Cu, n = 2 and Q = a(Zn²⁺)/a(Cu²⁺), approximately the concentration ratio in a dilute ideal model. More product relative to reactant reduces the forward voltage; pure solids are omitted because their activities are one.
For Ni+2Ag⁺→Ni²⁺+2Ag, n = 2 and Q = a(Ni²⁺)/a(Ag⁺)². For the reduction Mn⁺+ne⁻→M(s), the same equation gives E = E°+(RT/nF)ln a(Mn⁺). Powers in Q follow the balanced stoichiometric coefficients.
Go deeper: Keep temperature, pressure and units visible
Use kelvin in RT/nF. The 0.05916 V coefficient applies near 298 K and changes with temperature. Activities are dimensionless; concentration substitutions implicitly divide by the standard concentration. Gases contribute activity through pressure relative to standard pressure.
For 2H⁺+2e⁻⇌H₂ at 298 K and hydrogen pressure 1 bar, Q = 1/a(H⁺)², giving E = −0.05916 pH V relative to SHE. This relation is not an electrode-independent pH-to-voltage rule.
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Question
A Daniell cell at 298 K has a(Zn²⁺)=0.10 and a(Cu²⁺)=0.0010. Calculate E using E°=1.10 V and n=2.
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Equilibrium Constant and Gibbs Energy
Separate energy per charge from the amount of reaction.Voltage measures energy per charge
- Electrochemical Gibbs energy
- ΔrG = −nFE for the written reaction; standard values satisfy ΔrG° = −nFE° = −RT ln K.
A reversible cell can deliver a maximum electrical work whose magnitude is nFE per mole of the written reaction. A positive forward cell emf corresponds to negative ΔrG for that direction; a negative emf means an external driving source is needed. For Daniell E°=1.10 V and n=2, ΔrG° ≈ −212.27 kJ/mol reaction, using F≈96485 C/mol.
At equilibrium E=0, Q=K and ΔrG=0. Substitution into Nernst gives ln K=nFE°/RT or log₁₀K=nE°/0.05916 at 298 K. The Daniell example gives log₁₀K≈37.19 and K≈1.55×10³⁷ with these rounded inputs. The standard emf itself does not become zero merely because the reacting cell reaches equilibrium.
Go deeper: Scale the equation, not its voltage
For twice the written Daniell reaction, n doubles from 2 to 4 and ΔrG° doubles. E° stays 1.10 V because energy and charge scale together. An equilibrium constant changes to K² when every coefficient is doubled.
A thermodynamically favoured reaction may be slow because of activation barriers. Large K says where equilibrium lies, not how quickly it is approached. The energy expression refers to reversible emf, not an arbitrarily reduced terminal voltage under a heavy load.
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Question
A cell reaction has n=3 and E°=0.60 V. Find ΔrG° and state what changes when all reaction coefficients are doubled.
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Conductance and Conductivity
Keep κ, G, cell constant and Λm in separate steps.Separate the liquid from the measuring vessel
- Conductivity κ
- A material property equal to inverse resistivity; conductance G additionally depends on electrode geometry.
| Quantity | SI unit | Relation |
|---|---|---|
| Resistance R | Ω | R = ρl/A |
| Conductance G | S | G = 1/R = κA/l |
| Conductivity κ | S/m | κ = 1/ρ |
| Cell constant l/A | m⁻¹ | κR |
| Molar conductivity Λm | S m²/mol | κ / c, with c in mol/m³ |
Resistance R=ρl/A and conductance G=1/R=κA/l. R is measured in Ω, G in S, ρ in Ω m and κ in S/m. The cell constant l/A has units m⁻¹ and κ=(l/A)/R. It is normally calibrated with a solution of known conductivity rather than inferred from uncertain physical electrode dimensions.
Molar conductivity Λm=κ/c uses c in mol/m³ for units S m²/mol. Multiply molarity in mol/L by 1000 before using SI κ. Alternatively, κ in S/cm gives Λm=1000κ/c with c in mol/L and result S cm²/mol. 1 S m²/mol = 10⁴ S cm²/mol.
Go deeper: Ions and electrons carry current differently
Metallic conduction is mainly electronic and normally does not change bulk metal composition. Ionic conduction involves solvated cations and anions; nature, concentration, solvent viscosity, temperature and ion interactions influence conductivity. In many aqueous solutions ionic conductivity increases on warming, whereas metallic conductivity often falls.
A conductivity cell commonly uses platinised platinum electrodes and an AC measuring arrangement to reduce composition changes and polarisation. In a balanced bridge, Runknown=R1R4/R3 with the stated circuit labels. Prolonged DC passage can change electrode and solution composition, so measurement conditions matter.
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Question
A cell constant is 80 m⁻¹ and measured R is 400 Ω. For c=0.020 mol/L, find κ and Λm in both SI and cm² units.
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Variation of Conductivity and Kohlrausch Law
Diluting a solution decreases ions per volume but can improve conduction per mole.Dilution changes two different quantities
- Kohlrausch’s law
- At infinite dilution, limiting molar conductivity is the sum of stoichiometrically weighted individual ionic contributions.
As an electrolyte is diluted, conductivity κ typically decreases because fewer charge carriers occupy a given volume. Molar conductivity Λm increases because it measures conduction normalised to the amount of electrolyte. Strong electrolytes are largely dissociated; at low concentration, Λm≈Λ°m−A√c gives a near-linear plot against √c with intercept Λ°m.
Weak electrolytes dissociate more on dilution and their Λm rises steeply, so simple straight-line extrapolation is unsuitable. Kohlrausch’s law gives Λ°m=ν+λ°+ + ν−λ°−. Thus CaCl₂ includes two chloride contributions. For acetic acid, Λ°(HAc)=Λ°(HCl)+Λ°(NaAc)−Λ°(NaCl), cancelling sodium and chloride.
Go deeper: Relate weak-acid conduction to equilibrium carefully
For a dilute weak electrolyte whose ionic mobilities are approximated by limiting values, α≈Λm/Λ°m. For a monoprotic acid, Ka=cα²/(1−α). The model includes both the increasing degree of dissociation and dilution; concentration and molar conductivity must not be interchanged.
A straight strong-electrolyte relation is a dilute approximation, not an unrestricted law at high concentration. The coefficient A depends on solvent, temperature and electrolyte charge type. A solvent’s own conduction can matter as solute concentration becomes extremely small.
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Question
A weak acid has Λm=39.0 and Λ°m=390 S cm²/mol at c=0.0020 mol/L. Estimate α and Ka.
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Electrolysis and Faraday's Laws
Balance the electrode reaction before choosing n.Count electrons to count deposited atoms
- Faraday’s quantitative relation
- For constant current and 100% current efficiency, deposited mass m = MIt/(nF), with n electrons per deposited formula amount.
Faraday’s first law relates electrode reaction amount to passed charge Q; for constant current Q=It in coulombs when I is in amperes and t in seconds. One mole of electrons carries F≈96485 C. Faraday’s second law says the same charge produces masses proportional to chemical equivalent M/n, not equal masses.
Copper electrolysis with a reactive copper anode dissolves Cu(s)→Cu²⁺(aq)+2e⁻ at the positive anode and deposits Cu²⁺(aq)+2e⁻→Cu(s) at the negative cathode. This underlies copper refining: impure anode metal is transferred towards a purer cathode, while impurity behaviour depends on its chemistry.
Go deeper: Electrode material and water change the products
Molten NaCl produces Na at the cathode and Cl₂ at the anode. In concentrated aqueous brine with suitable electrodes, water reduction produces H₂ and OH⁻ while chloride oxidation produces Cl₂. Concentration, inert versus reactive electrodes and kinetic overpotential influence competition; standard potentials alone do not always identify the operating product.
Dilute aqueous sulphuric acid with suitable inert electrodes can give H₂ and O₂ from water. The chapter also contrasts sulphate oxidation under different conditions. Industrial extraction of reactive metals such as Na, Mg and Al uses appropriate molten electrolytes; the illustrated cathode stoichiometry does not prescribe an aqueous aluminium process.
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Question
A 2.0 A current passes for 15 minutes through a copper-plating cell. Find ideal Cu mass using M=63.546 g/mol, then explain why a measured mass could be lower.
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Batteries and Fuel Cells
Look inside the cell and identify what is actually reduced.Stored reactants or a continuous fuel supply
- Fuel cell
- A galvanic device supplied continuously with fuel and oxidant, with products removed during operation.
| Discharging cell | Oxidation side | Reduction side | Electrolyte |
|---|---|---|---|
| Dry cell | Zinc container | MnO₂ mixture; graphite collects current | Moist NH₄Cl/ZnCl₂ paste |
| Mercury-cell source example | Zn–Hg amalgam | HgO mixture | Alkaline paste |
| Lead storage | Pb plate | PbO₂ plate | Sulphuric acid solution |
| Alkaline H₂/O₂ fuel cell | H₂ at catalyst-bearing electrode | O₂ at catalyst-bearing electrode | Alkaline solution |
In a primary dry cell, the zinc casing is the anode and a graphite collector contacts a MnO₂-containing cathode mixture. Moist NH₄Cl/ZnCl₂ paste provides ionic conduction. The approximate cathode process MnO₂+NH₄⁺+e⁻→MnO(OH)+NH₃ reduces Mn from +4 to +3; nominal voltage is about 1.5 V. A primary cell is not designed for ordinary recharge.
The chapter’s mercury-cell example uses zinc amalgam and HgO with alkaline paste: Zn(Hg)+HgO→ZnO+Hg, near 1.35 V. The overall reaction does not consume changing solution-ion concentrations in that simplified model. Secondary lead storage cells discharge by Pb+PbO₂+2H₂SO₄→2PbSO₄+2H₂O; external charging reverses the chemistry. The nickel–cadmium cell is another rechargeable example with electrode layers separated by electrolyte-bearing material.
Go deeper: Compare stored chemicals with a feed stream
In an alkaline hydrogen–oxygen fuel cell, anode 2H₂+4OH⁻→4H₂O+4e⁻ and cathode O₂+2H₂O+4e⁻→4OH⁻ sum to 2H₂+O₂→2H₂O. OH⁻ travels internally from cathode towards anode while electrons pass externally through the load in the opposite direction. Porous catalyst-bearing electrodes allow gas supply and interfacial reaction.
A fuel cell can continue while reactants are supplied and suitable operating conditions remain. Its direct chemical-to-electrical conversion can be efficient, but no single efficiency fits every design or operating point. Water is the reaction product of this hydrogen cell; renewable supply and total environmental impact depend on hydrogen production, transport and equipment, not on the formula alone.
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Question
During lead-battery discharge both plates gain PbSO₄. Explain what charging must restore and whether a hydrogen fuel cell automatically uses renewable energy.
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Corrosion
Separate the anodic iron loss from the later formation of rust.Rust starts with separated oxidation and reduction
- Corrosion
- Deterioration of a material through chemical or electrochemical interaction with its surroundings; rusting is iron’s corrosion.
At an anodic spot on wet iron, Fe(s)→Fe²⁺(aq)+2e⁻. Electrons pass through the metal to a cathodic region where dissolved oxygen is reduced. In an aerated neutral-water model, O₂+2H₂O+4e⁻→4OH⁻; in an acidic model the corresponding reduction is O₂+4H⁺+4e⁻→2H₂O. The wet film carries ionic current.
The Fe²⁺ products undergo further oxidation and precipitation, eventually forming hydrated iron(III) oxide, often written Fe₂O₃·xH₂O. Thus the first anodic product is not already rust. Silver tarnishing and copper’s green surface products are other corrosion examples with different chemistry.
Go deeper: Barriers and sacrificial protection act differently
Painting or suitable surface treatments reduce contact with water and oxygen. A damaged barrier can expose the metal. Coating with another metal is not always simply an inert shield: active zinc can oxidise preferentially and supply protective electrons to electrically connected iron.
A sacrificial Zn or Mg anode is consumed while protecting the connected structure. Effectiveness depends on continued electrical contact, electrolyte access and remaining sacrificial material. A more noble tin coating can behave differently at a scratch, so coating identity matters as well as coverage.
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Question
Compare a small scratch in painted iron and a small scratch in zinc-coated iron under wet conditions. What physical path must remain for sacrificial protection?
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