12.1
Galvanic (Voltaic) Cells
Galvanic CellConverts chemical energy to electrical energy via a spontaneous redox reaction. Consists of two half-cells connected by a salt bridge.
Standard Electrode Potential E°
Standard hydrogen electrode (SHE): H⁺(1M) + e⁻ ⇋ ½H₂(g, 1atm) E° = 0.000 V (reference)
Cell EMF: E°cell = E°cathode − E°anode (reduction potentials)
(or: E°cell = E°(+) − E°(−))
Daniell cell: Zn|Zn²⁺(1M)||Cu²⁺(1M)|Cu
Zn → Zn²⁺ + 2e⁻ E° = +0.76 V (oxidation)... but listed as E°(Zn²⁺/Zn) = −0.76 V
Cu²⁺ + 2e⁻ → Cu E° = +0.34 V
E°cell = 0.34 − (−0.76) = +1.10 V
Positive E°cell → spontaneous; ΔG = −nFE°cell < 0
Key Standard Electrode Potentials
| Half-reaction | E° (V) |
|---|---|
| Li⁺ + e⁻ → Li | −3.04 |
| K⁺ + e⁻ → K | −2.93 |
| Na⁺ + e⁻ → Na | −2.71 |
| Al³⁺ + 3e⁻ → Al | −1.66 |
| Zn²⁺ + 2e⁻ → Zn | −0.76 |
| Fe²⁺ + 2e⁻ → Fe | −0.44 |
| 2H⁺ + 2e⁻ → H₂ | 0.00 |
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| Fe³⁺ + e⁻ → Fe²⁺ | +0.77 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.36 |
| F₂ + 2e⁻ → 2F⁻ | +2.87 |
12.2
Nernst Equation
Non-Standard Conditions
E = E° − (RT/nF)ln(Q)
At 25°C (298 K): E = E° − (0.0592/n)log(Q)
Where: n = moles of electrons transferred; Q = reaction quotient; F = 96,485 C/mol
At equilibrium: E = 0, Q = Kᵍᴺ
0 = E° − (0.0592/n)log(K)
log(K) = nE°/0.0592 → ΔG° = −nFE°
Example: Daniell cell with [Zn²⁺] = 2.0 mol/L, [Cu²⁺] = 0.01 mol/L
Q = [Zn²⁺]/[Cu²⁺] = 2.0/0.01 = 200
E = 1.10 − (0.0592/2)log(200) = 1.10 − 0.0296×2.301 = 1.10 − 0.068 = 1.03 V
12.3
Batteries and Cells
| Cell type | Anode | Cathode | Electrolyte | E (V) | Use |
|---|---|---|---|---|---|
| Leclanche (dry cell) | Zn | MnO₂/C | NH₄Cl paste | 1.5 | Torches, remotes |
| Alkaline dry cell | Zn | MnO₂/C | KOH paste | 1.5 | Higher capacity |
| Lead-acid (car battery) | Pb | PbO₂ | H₂SO₄(aq) | 2.0 per cell (6 cells=12V) | Car starter |
| Nickel-cadmium (NiCd) | Cd | NiO(OH) | KOH | 1.2 | Rechargeable tools |
| Lithium-ion | Graphite (Li intercalated) | LiCoO₂ | LiPF₆ in organic solvent | 3.7 | Phones, laptops, EVs |
| Hydrogen fuel cell | H₂ (Pt catalyst) | O₂ (Pt catalyst) | KOH or PEM | 1.23 | Vehicles, stationary |
Lead-Acid Battery
Anode (Pb): Pb + SO₄²⁻ → PbSO₄ + 2e⁻
Cathode (PbO₂): PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O (discharge)
Recharging reverses reaction (PbSO₄ → Pb + PbO₂).
H₂SO₄ is consumed on discharge — density of acid drops (measured by hydrometer).
12.4
Electrolytic Cells & Faraday’s Laws
Faraday’s Laws
1st Law: Mass deposited ∝ charge passed (Q = It)
2nd Law: Same charge deposits masses proportional to M/n (equivalent mass)
m = (M × I × t) / (n × F)
m = mass (g); M = molar mass; I = current (A); t = time (s); n = electrons; F = 96,485 C/mol
Example: Electroplate Ag (M=108) at 2.0 A for 30 min (1800 s), n=1:
m = (108 × 2.0 × 1800) / (1 × 96485) = 388,800/96,485 = 4.03 g Ag
12.5
Corrosion
Rusting of Iron
No videos added yet for this unit.
Electrochemical process: requires water + O₂ (+ electrolytes accelerate)
Anode (iron): Fe → Fe²⁺ + 2e⁻ (iron at pit, scratch)
Cathode (iron): O₂ + 2H₂O + 4e⁻ → 4OH⁻ (iron at O₂-rich surface)
Fe²⁺ + 2OH⁻ → Fe(OH)₂ → Fe(OH)₃ → Fe₂O₃·xH₂O (rust)
Prevention:
• Painting/coating (barrier)
• Galvanising (Zn sacrificial anode — Zn oxidises preferentially, protects Fe)
• Cathodic protection (attach Mg block; Mg anode, Fe cathodically protected)
• Stainless steel (Cr₂O₃ passive layer)
• Alloying, anodising, electroplating
✏️
Exercises
- Calculate E°cell for the cell: Zn|Zn²⁺||Fe³⁺,Fe²⁺|Pt. E°(Zn²⁺/Zn)=−0.76V; E°(Fe³⁺/Fe²⁺)=+0.77V.Cathode: Fe³⁺+e⁻→Fe²⁺ (+0.77V). Anode: Zn→Zn²⁺+2e⁻ (−0.76V). E°cell=0.77−(−0.76)=+1.53 V. Cell equation: Zn+2Fe³⁺→Zn²⁺+2Fe²⁺.
- Calculate the mass of aluminium deposited by passing 5.0 A for 2 hours through molten AlCl₃. Al M=27, n=3.Q=It=5.0×7200=36,000 C. m=MQ/(nF)=27×36000/(3×96485)=972,000/289,455=3.36 g Al.
- Using the Nernst equation, calculate the EMF of a cell Zn|Zn²⁺(0.001M)||Cu²⁺(0.1M)|Cu. E°cell=1.10V.Q=[Zn²⁺]/[Cu²⁺]=0.001/0.1=0.01. E=1.10−(0.0592/2)log(0.01)=1.10−0.0296×(−2)=1.10+0.059=1.16 V.
- Write equations for all steps in the rusting of iron. State the conditions needed.Anode (iron grain boundary or scratch): Fe→Fe²⁺+2e⁻. Cathode (oxygen-rich surface): O₂+2H₂O+4e⁻→4OH⁻. Then: Fe²⁺+2OH⁻→Fe(OH)₂; 4Fe(OH)₂+O₂→4Fe(OH)₃; 2Fe(OH)₃→Fe₂O₃·H₂O (rust) + water. Conditions required: water (electrolyte), O₂ (depolariser). Salt water greatly accelerates.
- Calculate Kᵍᴺ for the reaction Zn+Cu²⁺→Zn²⁺+Cu. E°cell=+1.10V, n=2.logK=nE°/0.0592=2×1.10/0.0592=37.16. K=10^37.16=1.45×10³⁷. Enormous K — reaction is essentially irreversible (products strongly favoured).
- Explain why Cu cannot reduce Fe²⁺ but Fe can reduce Cu²⁺.E°(Fe²⁺/Fe)=−0.44V; E°(Cu²⁺/Cu)=+0.34V. Fe is stronger reductant (more negative E°). Fe+Cu²⁺→Fe²⁺+Cu: E°cell=0.34−(−0.44)=+0.78V (spontaneous). Cu+Fe²⁺→Cu²⁺+Fe: E°cell=−0.44−0.34=−0.78V (non-spontaneous). Cu cannot reduce Fe²⁺ because Cu is less reactive.
🧠
Quiz — 25 Questions
Unit 12: Electrochemical Cells & Applications
25 QsQ1
In a galvanic cell, oxidation occurs at the:
Anode = oxidation (loss of electrons). Cathode = reduction (gain of electrons). Mnemonic: RED CAT (Reduction At Cathode), AN OX (ANode OXidation).
Q2
E°cell = E°cathode − E°anode. For the Daniell cell (Zn/Cu):
Cu²⁺/Cu: E°=+0.34 V (cathode). Zn²⁺/Zn: E°=−0.76 V (anode). E°cell=0.34−(−0.76)=+1.10 V. Positive → spontaneous.
Q3
ΔG° = −nFE°cell. For a cell with E°=+0.50V and n=2:
ΔG°=−nFE°=−2×96485×0.50=−96,485 J/mol = −96.5 kJ/mol. Negative → spontaneous. ΔG° and E° are linked: positive E° ↔ negative ΔG° ↔ spontaneous reaction.
Q4
The Nernst equation is E = E° − (0.0592/n)logQ. At equilibrium:
At equilibrium: net cell voltage = 0 (no driving force). E=0, Q=K. From Nernst: 0=E°−(0.0592/n)logK → logK=nE°/0.0592. This allows K to be calculated from E°.
Q5
In a lead-acid battery during discharge:
Anode: Pb+SO₄²⁻→PbSO₄+2e⁻; Cathode: PbO₂+SO₄²⁻+4H⁺+2e⁻→PbSO₄+2H₂O. Both electrodes coated with PbSO₄; H₂SO₄ consumed → density drops. Recharging reverses.
Q6
Faraday's first law of electrolysis states:
1st law: m ∝ Q = It. 2nd law: same charge deposits masses in ratio of M/n (equivalent masses). These are empirical laws confirmed by experiment and explained by ionic theory.
Q7
Calculate mass of Cu (M=63.5) deposited by 1.5 A for 10 minutes (n=2):
m=MIt/(nF)=63.5×1.5×600/(2×96485)=57150/192970=0.296 g.
Q8
A hydrogen fuel cell produces electricity from:
H₂ fuel cell: H₂→2H⁺+2e⁻ (anode); O₂+4H⁺+4e⁻→2H₂O (cathode). Overall: 2H₂+O₂→2H₂O + electricity + heat. No combustion — direct conversion, higher efficiency (~60% vs ~35% for heat engine).
Q9
Galvanising protects steel by:
Zn (E°=−0.76V) is more reactive than Fe (E°=−0.44V). Even when Zn coating is scratched: Zn→Zn²⁺+2e⁻ (anodic), electrons flow to Fe (cathodic, protected). Fe does not oxidise while Zn is present. Only when Zn is fully consumed does Fe corrode.
Q10
The standard hydrogen electrode (SHE) has E° = 0.000 V because:
E° values are relative measurements. The SHE (H⁺(1M) + e⁻ ⇋ ½H₂, 1 atm Pt) is chosen as the reference with E°=0.000 V by convention. All other E° values are measured against it.
Q11
In electrolysis, at the cathode:
Electrolytic cathode = reduction (same as galvanic cathode). Cations (positive ions) migrate to cathode and gain electrons: Mⁿ⁺ + ne⁻ → M. This is the basis of electroplating, electrorefining, and metal extraction (e.g. Cu electrorefining).
Q12
Cathodic protection of underground steel pipelines uses:
Mg (E°=−2.37V) or Zn (E°=−0.76V) blocks attached to the pipeline: Mg oxidises preferentially → electrons flow to pipeline → pipeline is cathode → iron is not oxidised. Also: impressed current (rectifier forces pipeline negative = cathode). Used for pipelines, ship hulls, harbour structures.
Q13
E°(Ag⁺/Ag)=+0.80V; E°(Cu²⁺/Cu)=+0.34V. Silver can be deposited from Cu|Ag cell. The cathode reaction is:
Ag⁺/Ag (+0.80V) is reduced; Cu/Cu²⁺ (−0.34V when reversed) is oxidised. Cathode: Ag⁺+e⁻→Ag. Anode: Cu→Cu²⁺+2e⁻. E°cell=0.80−0.34=+0.46V (spontaneous).
Q14
Increasing concentration of the reducing agent in a galvanic cell:
Nernst: E=E°−(0.0592/n)logQ. If [reductant] increases → Q decreases → logQ decreases → E increases. Concentration cell: two identical electrodes at different concentrations — the higher concentration side is cathode.
Q15
The salt bridge in a galvanic cell serves to:
Without salt bridge: anode half-cell accumulates positive charge (metal ions dissolve), cathode accumulates negative. This charge imbalance stops current. Salt bridge (KCl or KNO₃ in agar) allows ions to migrate: K⁺ into cathode, Cl⁻ into anode → neutrality maintained → current flows.
Q16
Which cell reaction has the largest E°cell?
Li (E°=−3.04V) anode; F₂ (E°=+2.87V) cathode. E°cell=2.87−(−3.04)=+5.91V. Li/F₂ has the highest possible cell voltage with common elements. Not practical as a battery (F₂ too reactive), but shows the principle.
Q17
Electrorefining of copper:
Anode: impure Cu dissolves (Cu→Cu²⁺+2e⁻). Cathode: pure Cu deposits (Cu²⁺+2e⁻→Cu). Less reactive impurities (Au, Ag, Pt) do not dissolve — fall as anode slime (valuable by-product). More reactive impurities (Fe, Zn, Ni) dissolve but don’t plate at cathode voltage used. Gives 99.999% pure Cu.
Q18
A cell runs at 0.50 A. How much time is needed to deposit 1.0 g of Ni (M=58.7, n=2)?
m=MIt/(nF) → t=mnF/(MI)=1.0×2×96485/(58.7×0.50)=192970/29.35=6,577 s ≈ 6570 s (≈1.83 hours).
Q19
The EMF of a concentration cell Cu|Cu²⁺(0.01M)||Cu²⁺(1.0M)|Cu is:
E°=0 (identical electrodes). Q=[Cu²⁺]anode/[Cu²⁺]cathode=0.01/1.0=0.01. E=0−(0.0592/2)log(0.01)=−0.0296×(−2)=+0.059 V. Current flows from dilute to concentrated side (dilute is anode).
Q20
Rusting is fastest when:
Salt water: high ionic strength → high conductivity → easy ion transport between anodic and cathodic areas → large corrosion current. Fe is the anode at scratches/pits (different oxygen concentrations). O₂ acts as cathode depolariser. Pure water corrodes slowly (low conductivity); dry air: no electrolyte, very slow.
Q21
The relationship between K and E° is:
From Nernst at equilibrium: 0=E°−(0.0592/n)logK → logK = nE°/0.0592 (at 25°C). E.g. Daniell cell E°=1.10V, n=2: logK=2×1.10/0.0592=37.2; K=10^37.2=huge (reaction essentially complete).
Q22
In the alkaline fuel cell, the cathode reaction is:
Cathode (reduction in alkaline: KOH electrolyte): O₂+2H₂O+4e⁻→4OH⁻. Anode: H₂+2OH⁻→2H₂O+2e⁻. Overall: 2H₂+O₂→2H₂O. Used in Apollo spacecraft. In PEM fuel cell (acid): cathode: O₂+4H⁺+4e⁻→2H₂O.
Q23
Lithium-ion batteries have E≈3.7V because:
Li (E°=−3.04V) combined with LiCoO₂ (E°≈+0.6V) gives ≈3.7V. Li⁺ intercalates (not pure Li, so safer). Very high energy density (~250 Wh/kg) due to light mass of Li and high voltage. Dominant technology for portable electronics and EVs.
Q24
Which metal provides cathodic protection to ship hulls?
Zn or Mg blocks (sacrificial anodes) attached to steel ship hull. Mg/Zn are more reactive (more negative E°) → they oxidise preferentially, giving electrons to hull (cathodic). Hull protected. Blocks replaced periodically when consumed.
Q25
During electrolysis of dilute H₂SO₄, the products at each electrode are:
Cathode: 2H⁺+2e⁻→H₂ (reduction). Anode: 2H₂O→O₂+4H⁺+4e⁻ (oxidation of water, since SO₄²⁻ is hard to oxidise). Overall: 2H₂O→2H₂+O₂ (electrolysis of water). H₂SO₄ concentration increases as water is consumed.
Unit 12 Quiz — Electrochemical Cells (25 Questions)
Select one answer eachQ1
In a galvanic (voltaic) cell, oxidation occurs at the:
Anode = oxidation (OIL — Oxidation Is Loss). The more reactive metal loses electrons at the anode.
Q2
The standard electrode potential (E°) is measured relative to:
SHE: Pt | H₂(g, 1 atm) | H⁺(aq, 1 mol/L), E° = 0.00 V by convention. All other E° measured against this.
Q3
The standard cell potential E°cell is calculated as:
E°cell = E°cathode – E°anode (reduction potential of cathode minus reduction potential of anode). Positive E°cell = spontaneous.
Q4
If E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = –0.76 V, the E°cell for a Zn-Cu cell is:
E°cell = E°cathode – E°anode = +0.34 – (–0.76) = +1.10 V. Cu²⁺ is reduced (cathode); Zn is oxidised (anode).
Q5
The Nernst equation accounts for:
Nernst: E = E° – (RT/nF)lnQ. At non-standard concentrations or temperatures, E differs from E°.
Q6
The Nernst equation at 25°C simplifies to:
At 298 K: RT/F = 0.02569 V. With log base 10: E = E° – (0.05916/n)logQ ≈ E° – (0.0592/n)logQ.
Q7
At equilibrium, a galvanic cell has:
When Q = K, ΔG = 0 and E = 0. The cell has run down — no more driving force. Relationship: lnK = nFE°/RT.
Q8
The salt bridge in a galvanic cell:
As charge builds up (anode +, cathode –), ion flow through salt bridge neutralises charges — maintains circuit.
Q9
Electrolytic cells differ from galvanic cells in that:
Galvanic: spontaneous → electricity. Electrolytic: non-spontaneous → requires external EMF to force reaction.
Q10
During electrolysis of aqueous CuSO₄ with copper electrodes:
Cathode: Cu²⁺ + 2e⁻ → Cu. Anode: Cu → Cu²⁺ + 2e⁻. [CuSO₄] stays constant — used in copper refining.
Q11
Standard Gibbs energy change is related to E°cell by:
ΔG° = –nFE°cell. If E°cell > 0, ΔG° < 0 — spontaneous. n = moles of electrons; F = Faraday constant (96485 C/mol).
Q12
The Faraday constant (F) represents:
F = eNA = 1.602×10⁻¹⁹ × 6.022×10²³ = 96485 C mol⁻¹. Used in electrochemical calculations.
Q13
Mass of metal deposited in electrolysis is given by Faraday's law:
Moles of ion = Q/(nF) = It/(nF). Mass = moles × M(molar mass). E.g. Cu²⁺: n=2; 1 mol Cu needs 2F.
Q14
A fuel cell produces electricity by:
H₂–O₂ fuel cell: H₂ → 2H⁺ + 2e⁻ (anode), O₂ + 4H⁺ + 4e⁻ → 2H₂O (cathode). No combustion; ~60% efficient.
Q15
The lithium-ion battery uses lithium because:
Li/Li⁺: E° = –3.04 V — lowest in the electrochemical series. High energy density per unit mass.
Q16
In the chlor-alkali process, electrolysis of brine gives:
Anode: 2Cl⁻ → Cl₂ + 2e⁻. Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻. NaOH remains in solution.
Q17
Cathodic protection prevents iron from rusting by:
Connected to a more reactive metal (Mg, Zn) as anode — these corrode sacrificially while iron stays reduced (cathode).
Q18
The electrochemical series arranges metals in order of:
More negative E° = more easily oxidised (stronger reducing agent). Predicts which metal displaces another.
Q19
A concentration cell has:
Both electrodes same metal (e.g. Cu); different [Cu²⁺] in each half. E = (0.0592/2)log([Cu²⁺]dilute/[Cu²⁺]concentrated). Spontaneous.
Q20
Which electrode reaction occurs at the cathode of a hydrogen fuel cell?
Cathode (reduction): O₂ + 4H⁺ + 4e⁻ → 2H₂O. Anode (oxidation): H₂ → 2H⁺ + 2e⁻. Net: 2H₂ + O₂ → 2H₂O.
Q21
lnK = nFE°/RT implies that a cell with E°cell = 0 has:
If E° = 0, then lnK = 0 → K = 1. This means ΔG° = 0 — products and reactants equally favoured at standard conditions.
Q22
Electroplating silver onto a spoon requires the spoon to be the:
Cathode: Ag⁺ + e⁻ → Ag(s) deposited on spoon. Silver anode dissolves to replenish Ag⁺ in solution.
Q23
The lead-acid battery (car battery) is a:
Discharge: Pb → Pb²⁺ (anode), PbO₂ → Pb²⁺ (cathode). Both give PbSO₄. Recharging reverses the reactions.
Q24
Overpotential in electrolysis is:
Practical electrolysis needs more voltage than E°cell predicts — overpotential compensates for slow electrode reactions and solution resistance.
Q25
Daniel cell uses:
Classic galvanic cell: Zn → Zn²⁺ + 2e⁻ (anode), Cu²⁺ + 2e⁻ → Cu (cathode). E°cell = 0.34–(–0.76) = 1.10 V.
📝
Unit Test — 50 marks
Section A
30 marksQ1 [5]
For the cell: Fe|Fe²⁺(1M)||MnO₄⁻,Mn²⁺,H⁺|Pt. E°(Fe²⁺/Fe)=−0.44V; E°(MnO₄⁻/Mn²⁺)=+1.51V. (a) Write electrode equations. (b) Calculate E°cell. (c) Calculate ΔG°. (d) Is the reaction spontaneous? [5]
(a) Anode: Fe→Fe²⁺+2e⁻; Cathode: MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O. (b) E°cell=1.51−(−0.44)=+1.95V. (c) To balance electrons: 5Fe→5Fe²⁺+10e⁻; 2MnO₄⁻+16H⁺+10e⁻→2Mn²⁺+8H₂O; n=10. ΔG°=−nFE°=−10×96485×1.95=−1,881,458 J=−1881 kJ/mol. (d) Spontaneous (E°>0, ΔG°<0).
Q2 [5]
An electrolytic cell plates nickel (M=58.7, n=2) onto steel at 3.0A. (a) Calculate the mass deposited in 1 hour. (b) Calculate the thickness if the area is 200 cm² and density of Ni = 8.9 g/cm³. [5]
(a) m=MIt/(nF)=58.7×3.0×3600/(2×96485)=633,960/192,970=3.29g. (b) Volume=mass/density=3.29/8.9=0.370 cm³. Thickness=volume/area=0.370/200=0.00185cm=18.5 μm.
Q3 [5]
Describe the lead-acid battery: half-cell reactions, overall reaction, why density of H₂SO₄ decreases on discharge, and how recharging works. [5]
Discharge: Anode(Pb): Pb+SO₄²⁻→PbSO₄+2e⁻; Cathode(PbO₂): PbO₂+SO₄²⁻+4H⁺+2e⁻→PbSO₄+2H₂O. Overall: Pb+PbO₂+2H₂SO₄→2PbSO₄+2H₂O. H₂SO₄ consumed → [H₂SO₄] decreases → density of solution drops. Measured by hydrometer to test state of charge. Recharging: external power supply reverses reactions: PbSO₄→Pb (anode, now cathode) and PbSO₄→PbO₂ (cathode, now anode); H₂SO₄ regenerated.
Q4 [5]
Calculate the equilibrium constant K for: 2Ag⁺(aq)+Cu(s)⇋2Ag(s)+Cu²⁺(aq). E°(Ag⁺/Ag)=+0.80V; E°(Cu²⁺/Cu)=+0.34V. [5]
Cathode: 2Ag⁺+2e⁻→2Ag (+0.80V). Anode: Cu→Cu²⁺+2e⁻ (−0.34V). E°cell=0.80−0.34=+0.46V; n=2. logK=nE°/0.0592=2×0.46/0.0592=15.54. K=10^15.54=3.5×10^15. Essentially irreversible (Cu displaces Ag⁺ spontaneously).
Q5 [5]
Explain three methods of corrosion protection and the electrochemical principle underlying each. [5]
1. Galvanising (Zn coat on Fe): Zn (E°=−0.76V) < Fe (E°=−0.44V). Even at scratches: Zn oxidises preferentially (sacrificial anode), Fe remains cathodically protected. 2. Cathodic protection (Mg/Zn blocks on pipeline): Mg (E°=−2.37V) more reactive: Mg→Mg²⁺+2e⁻; electrons flow to pipe; pipe is cathode (Fe²⁺+2e⁻→Fe), not oxidised. 3. Painting/coating: physical barrier prevents water+O₂ contact — both are required for electrochemical corrosion. No electrolyte → no ionic transport → no cell current → no corrosion. 4. Stainless steel: Cr (16–18%) forms passivating Cr₂O₃ layer on surface — prevents further oxidation. Self-healing: if scratched, Cr₂O₃ reforms in presence of O₂.
Q6 [5]
The Nernst equation predicts that concentration cells generate EMF. (a) Explain this with a Cu concentration cell: Cu|Cu²⁺(0.001M)||Cu²⁺(1.0M)|Cu. Calculate E. (b) At what concentration ratio does E reach 0.10V (n=2)?
(a) E°=0 (same electrodes). Q=0.001/1.0=10⁻³. E=0−(0.0592/2)log(10⁻³)=−0.0296×(−3)=+0.089V. Anode: dilute (0.001M); cathode: concentrated (1.0M). Electrons flow from anode to cathode — concentration difference drives current. (b) 0.10=0−(0.0592/2)logQ → logQ=−0.10/0.0296=−3.38 → Q=10⁻³·³⁸=4.17×10⁻⁴. So ratio [dilute]/[concentrated]=4.2×10⁻⁴ (e.g. 0.0001M vs 0.24M).
Section B
20 marksQ7 [10]
(a) Describe the standard hydrogen electrode. Why is it used as the reference electrode? [3] (b) Explain how E° values are used to predict the feasibility of redox reactions and calculate ΔG° and K. Use Fe/Cu as an example. [7]
(a) SHE: Pt wire electrode in 1M H⁺(aq) with H₂ gas at 1 atm bubbling over it. Pt is inert (doesn’t react). Half-reaction: H⁺(1M)+e⁻⇋½H₂(1 atm). E°=0.000V by international convention. Used as reference because: (1) reproducible (any lab can make it); (2) covers the middle of the E° range; (3) defined as zero providing a consistent reference point. All other E° values are measured vs SHE under standard conditions (1M concentrations, 1 atm, 25°C).
(b) Predicting feasibility: E°cell=E°cathode−E°anode. If E°cell>0 → spontaneous. For Fe+Cu²⁺→Fe²⁺+Cu: cathode Cu²⁺/Cu E°=+0.34V; anode Fe/Fe²⁺ E°=−0.44V. E°cell=0.34−(−0.44)=+0.78V (spontaneous). ΔG°=−nFE°=−2×96485×0.78=−150,517J=−150.5kJ/mol (negative → spontaneous). logK=nE°/0.0592=2×0.78/0.0592=26.35; K=10^26.35=2.2×10^26 (reaction essentially irreversible, strongly to completion). Reverse reaction (Cu reducing Fe²⁺): E°cell=−0.78V (negative → non-spontaneous). This confirms the activity series: Fe is above Cu — Fe displaces Cu²⁺ from solution.
Q8 [10]
Compare galvanic cells and electrolytic cells. Discuss lithium-ion batteries and hydrogen fuel cells as modern electrochemical technology, including their advantages, disadvantages, and environmental impact. [10]
Galvanic vs electrolytic comparison: Galvanic cell: spontaneous reaction (ΔG<0), converts chemical to electrical energy, anode(−) is negative pole (oxidation, electrons flow out), cathode(+) is positive. Examples: batteries, fuel cells. Electrolytic cell: non-spontaneous reaction driven by external power (ΔG>0), converts electrical to chemical energy, anode(+) connected to positive terminal. Examples: electroplating, electrolysis, aluminium extraction.
Lithium-ion batteries: Li⁺ intercalates in graphite anode on charge. Cathode: LiCoO₂ (or LiFePO₄, NMC). E≈3.7V, high energy density (250 Wh/kg), long cycle life (500–2000 cycles), low self-discharge. Advantages: high power-to-weight ratio (ideal for EVs, portable devices), fast charge. Disadvantages: Co is expensive and conflict-mineral; thermal runaway risk (fire if damaged); limited by lithium supply; capacity degrades over cycles; performance drops in cold. Environmental: mining Li and Co (Chile, DRC) has environmental/social impacts; however, over lifetime, EV with Li-ion battery emits far less CO₂ than ICE vehicle even accounting for manufacturing.
Hydrogen fuel cells: H₂ oxidised at anode (H₂→2H⁺+2e⁻); O₂ reduced at cathode (O₂+4H⁺+4e⁻→2H₂O). Only waste = water. E°=1.23V; actual efficiency ~50–60% (vs ~35% internal combustion). PEM (proton exchange membrane) cells for transport. Advantages: zero tailpipe emissions, quick refuelling (5 min vs hours to charge battery), high energy density per kg of H₂. Disadvantages: H₂ storage difficult (cryogenic liquid at −253°C or compressed at 700 bar); Pt catalyst expensive and scarce; 95% of H₂ currently made from natural gas (grey H₂, produces CO₂); green H₂ (electrolysis using renewable power) still expensive. Infrastructure lacking (few refuelling stations). Environmental: green H₂ + fuel cell = truly zero carbon transport; grey H₂ = similar GHG to conventional fuels. Future potential high, especially for heavy transport (trucks, trains, ships) where batteries are too heavy.