Galvanic cells, electrode potential and the electrochemical series
Link to Std 11 : In Std 11 you learnt redox reactions and the activity series of metals; a galvanic cell uses that electron transfer to produce current.
A torch cell turns a chemical reaction into an electric current that lights the bulb.
A galvanic cell converts the chemical energy of a spontaneous redox reaction into electrical energy. Oxidation happens at the anode (negative) and reduction at the cathode (positive). The standard electrode potential E° is the potential of an electrode with all species at 1 mol L⁻¹ (gases at 1 bar), measured against the standard hydrogen electrode, which is fixed at 0 V.
Galvanic cells, electrode potential and the electrochemical series
In the app you can move this figure with your finger.
Open in the appIn the figure : Electrons are released at the zinc anode, travel through the outer wire to the copper cathode, and the salt bridge completes the circuit by letting ions move so that each half-cell stays electrically neutral.
Rule 1: Cell notation and E°cell — Write the anode on the left and the cathode on the right: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), where | is a phase boundary and || is the salt bridge. E°cell = E°(cathode) − E°(anode), using reduction potentials for both.
- Easy : Daniell cell: E°cell = 0.34 − (−0.76) = 1.1 V.
- Medium : Cu | Cu²⁺ || Ag⁺ | Ag has E°cell = 0.80 − 0.34 = 0.46 V.
- Exam-type : Mg | Mg²⁺ || Ag⁺ | Ag has E°cell = 0.80 − (−2.36) = 3.16 V; the half-cell with the more negative E° is the anode.
Rule 2: Electrochemical series — Arranging metals by E° gives the electrochemical series. A species with a more negative E° is a stronger reducing agent; one with a more positive E° is a stronger oxidising agent. A metal displaces from solution any metal with a more positive E°. Values (V): Li −3.05, K −2.93, Na −2.71, Mg −2.36, Al −1.66, Zn −0.76, Fe −0.44, Pb −0.13, H₂ 0.00, Cu +0.34, Ag +0.80.
- Easy : Zinc can displace copper from copper sulphate solution.
- Medium : Copper cannot displace zinc from zinc sulphate, so a copper vessel can hold ZnSO₄ solution but a zinc vessel cannot hold CuSO₄ solution.
- Exam-type : Which is the stronger reducing agent, Fe (−0.44 V) or Zn (−0.76 V)? Zn, because its E° is more negative.
Rule 3: Spontaneity — A cell reaction is spontaneous when E°cell is positive. If a calculated E°cell is negative, the reverse reaction is the spontaneous one.
- Easy : Cu + Zn²⁺ → Cu²⁺ + Zn: E°cell = −0.76 − 0.34 = −1.10 V, so the reaction does not occur on its own.
- Medium : Fe²⁺ + Zn → Fe + Zn²⁺: E°cell = −0.44 − (−0.76) = +0.32 V, so it is spontaneous.
- Exam-type : A cell uses Pb²⁺/Pb (−0.13 V) and Al³⁺/Al (−1.66 V). Al is the anode, E°cell = −0.13 − (−1.66) = 1.53 V.
The more negative half-cell E° is the anode; E°cell is always cathode minus anode, so it is positive for a working cell.
Multiplying E° by the number of electrons while balancing the equation. E° is an intensive value and is NOT multiplied.
Memory clue: "AN OX, RED CAT" — ANode = OXidation, REDuction = CAThode; and E°cell = cathode minus anode.
Spot the difference
| Point | Anode | Cathode |
|---|---|---|
| Process | Oxidation | Reduction |
| Sign in galvanic cell | Negative | Positive |
| Electrons | Released | Gained |
Practice questions
- 1Which electrode is the anode in a galvanic cell?
- 2State the E° value assigned to the standard hydrogen electrode.
- 3What is the use of a salt bridge?
- 4Find E°cell of Cu | Cu²⁺ || Ag⁺ | Ag (E° Cu = 0.34 V, Ag = 0.80 V).
- 5Can an iron spoon stir a solution of ZnSO₄ without reaction? (Fe −0.44 V, Zn −0.76 V)
- 6Write the cell notation for the reaction Zn + Cu²⁺ → Zn²⁺ + Cu.
- 7A cell is made from Al³⁺/Al (−1.66 V) and Pb²⁺/Pb (−0.13 V). Identify anode, cathode and E°cell.
- 8Why can zinc be used to reduce Cu²⁺ ions but silver cannot reduce Zn²⁺ ions?
Show answers (try them yourself first)
- 1The electrode where oxidation takes place (negative terminal in a galvanic cell).
- 20.00 V.
- 3It completes the electrical circuit and keeps both solutions electrically neutral by allowing ion movement.
- 4E°cell = 0.80 − 0.34 = 0.46 V.
- 5Yes, no reaction: Fe has a higher E° than Zn, so it cannot reduce Zn²⁺.
- 6Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).
- 7Anode: Al; cathode: Pb; E°cell = −0.13 − (−1.66) = 1.53 V.
- 8Zn (E° −0.76 V) is a stronger reducing agent than Cu (+0.34 V), so it donates electrons to Cu²⁺; silver (E° +0.80 V) is a far weaker reducing agent than Zn.
Nernst equation, Gibbs energy and equilibrium constant
Link to Std 11 : Std 11 gave you ΔG = ΔH − TΔS and the reaction quotient Q; here they are tied to the voltage of a cell.
A torch battery goes dim as it is used because the reaction slowly moves towards equilibrium, where the voltage is zero.
The Nernst equation gives the cell potential at any concentrations: E = E° − (2.303 RT ÷ nF) log Q, which at 298 K is E = E° − (0.0592 ÷ n) log Q. At equilibrium E = 0 and Q = K. Also ΔrG° = −nFE° and ΔrG° = −RT ln K.
Nernst equation, Gibbs energy and equilibrium constant
In the app you can move this figure with your finger.
Open in the appIn the figure : The line starts at E° when Q is very small and falls with a slope of 0.0592 ÷ n; when Q reaches K the cell is flat (dead).
Rule 1: Using the Nernst equation — For Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺] ÷ [Cu²⁺]. Pure solids are left out of Q. n is the number of electrons transferred in the balanced equation.
- Easy : [Zn²⁺] = 0.01 M and [Cu²⁺] = 1.0 M: Q = 0.01, E = 1.10 − (0.0592 ÷ 2) × log 0.01 = 1.10 + 0.0592 = 1.159 V.
- Medium : [Zn²⁺] = 0.50 M and [Cu²⁺] = 0.002 M: Q = 250, E = 1.10 − 0.0296 × 2.398 = 1.029 V.
- Exam-type : A concentration cell with the same metal in both half-cells has E° = 0, so E = −(0.0592 ÷ n) log Q; for Cu²⁺ at 0.01 M and 1.0 M, E = 0.0296 × 2 = 0.0592 V.
Rule 2: Gibbs energy of the cell reaction — ΔrG° = −nFE°cell, with F = 96500 C mol⁻¹. Work obtained from the cell in one mole of reaction is −ΔG. A positive E° gives a negative ΔG (spontaneous).
- Easy : Daniell cell: ΔrG° = −2 × 96500 × 1.10 = -212.3 kJ mol⁻¹.
- Medium : If E°cell = −0.20 V with n = 2, ΔrG° = +38.6 kJ mol⁻¹ (non-spontaneous).
- Exam-type : For a cell with E°cell = 0.46 V and n = 2: ΔrG° = −2 × 96500 × 0.46 = -88.8 kJ mol⁻¹.
Rule 3: Equilibrium constant — From ΔrG° = −RT ln K and ΔrG° = −nFE°: log K = nE° ÷ 0.0592 at 298 K. A large positive E° means a large K and a reaction that goes almost to completion.
- Easy : Daniell cell: log K = 2 × 1.10 ÷ 0.0592 = 37.16, so K is about 1.4 × 10³⁷.
- Medium : E° = 0 gives log K = 0, so K = 1.
- Exam-type : A cell with E° = 0.0592 V and n = 1: log K = 1, K = 10.
E° is for 1 M and 1 bar; E uses actual concentrations. At equilibrium E = 0 and the battery is dead.
Putting Q = products ÷ reactants the wrong way round, which flips the sign of the correction.
Memory clue: "Q small, E high" — a cell with little product has E above E°; as products pile up, E falls to zero.
Spot the difference
| Point | E°cell | Ecell |
|---|---|---|
| Concentration | All 1 M | Any |
| Equation | E°cathode − E°anode | E° − (0.0592 ÷ n) log Q |
| At equilibrium | Not zero | Zero |
Practice questions
- 1What is the value of E at equilibrium?
- 2Write ΔrG° in terms of E°cell.
- 3Write the Nernst equation for 298 K.
- 4Find ΔrG° for a cell with E° = 0.46 V and n = 2.
- 5Find E for the Daniell cell when [Zn²⁺] = 0.01 M and [Cu²⁺] = 1.0 M.
- 6Find log K for a cell with E° = 0.30 V and n = 2.
- 7Find K for the Daniell cell (E° = 1.10 V, n = 2) at 298 K.
- 8For Cu²⁺ (0.01 M) | Cu || Cu²⁺ (1.0 M) | Cu, find the emf (n = 2).
Show answers (try them yourself first)
- 1Zero.
- 2ΔrG° = −nFE°cell.
- 3E = E° − (0.0592 ÷ n) log Q.
- 4ΔrG° = −2 × 96500 × 0.46 = -88.8 kJ mol⁻¹.
- 5E = 1.10 − 0.0296 × log(0.01) = 1.10 + 0.0592 = 1.159 V.
- 6log K = 2 × 0.30 ÷ 0.0592 = 10.14.
- 7log K = 2 × 1.10 ÷ 0.0592 = 37.16; K = 1.4 × 10³⁷.
- 8E = 0 − (0.0592 ÷ 2) × log(0.01 ÷ 1.0) = +0.0592 V.
Conductance of solutions and Kohlrausch's law
Link to Std 11 : In Std 11 you met strong and weak electrolytes and ionisation; conductivity measures how well the ions carry current.
Pure water hardly conducts, but when salt is added the bulb lights because ions are present to carry the current.
Conductivity κ is the conductance of a solution between electrodes 1 cm apart of area 1 cm². Molar conductivity Λm = κ × 1000 ÷ c (κ in S cm⁻¹, c in mol L⁻¹, Λm in S cm² mol⁻¹). Kohlrausch's law: the limiting molar conductivity Λ°m of an electrolyte is the sum of the contributions of its ions, Λ°m = ν₊λ°₊ + ν₋λ°₋.
Conductance of solutions and Kohlrausch's law
In the app you can move this figure with your finger.
Open in the appIn the figure : On dilution κ falls (fewer ions per cm³) but Λm rises (the ions move more freely and, for weak electrolytes, more of the acid ionises). The strong electrolyte line is nearly straight and meets the axis at Λ°m; the weak electrolyte curve cannot be extrapolated.
Rule 1: Resistance, cell constant and conductivity — R = ρ l ÷ A, G = 1 ÷ R (siemens, S), κ = G × (l ÷ A) = G × G*, where G* = l ÷ A is the cell constant (cm⁻¹). The cell constant is found from a solution of known conductivity such as KCl.
- Easy : A cell filled with 0.10 M KCl (κ = 0.0129 S cm⁻¹) has R = 100 Ω, so G* = κ × R = 0.0129 × 100 = 1.29 cm⁻¹.
- Medium : The same cell filled with a sample of R = 250 Ω has κ = G* ÷ R = 1.29 ÷ 250 = 0.00516 S cm⁻¹.
- Exam-type : A.C. (not D.C.) is used for measurement so that electrolysis does not change the solution.
Rule 2: Molar conductivity and dilution — Λm = κ × 1000 ÷ c. For strong electrolytes Λm rises slowly on dilution; Λm = Λ°m − A √c (Debye–Hückel–Onsager). For weak electrolytes it rises very steeply near infinite dilution.
- Easy : 0.10 M KCl with κ = 0.0129 S cm⁻¹: Λm = 0.0129 × 1000 ÷ 0.10 = 129 S cm² mol⁻¹.
- Medium : On diluting a solution, conductivity κ decreases because there are fewer ions per unit volume.
- Exam-type : Why does Λm increase on dilution? The total volume that contains one mole becomes larger, ions are farther apart and interionic attraction is less (and weak electrolytes ionise more).
Rule 3: Kohlrausch's law and its uses — Ions move independently at infinite dilution, so Λ°m(electrolyte) = λ°(cation) + λ°(anion). This lets us find Λ°m of weak electrolytes, and the degree of dissociation α = Λm ÷ Λ°m, and then Ka = c α² ÷ (1 − α).
- Easy : Acetic acid: λ°(H⁺) = 349.6 and λ°(CH₃COO⁻) = 40.9, so Λ°m = 390.5 S cm² mol⁻¹.
- Medium : A 0.05 M acetic acid has Λm = 7.8, so α = 7.8 ÷ 390.5 = 0.02.
- Exam-type : Ka = cα² ÷ (1 − α) = 0.05 × (0.02)² ÷ 0.98 = 2.0 × 10⁻⁵.
Conductivity falls and molar conductivity rises when a solution is diluted.
Using c in mol m⁻³ with κ in S cm⁻¹ without changing units. With c in mol L⁻¹ and κ in S cm⁻¹, the factor 1000 is needed.
Memory clue: "κ down, Λ up" — on dilution kappa falls, lambda rises.
Spot the difference
| Point | Strong electrolyte | Weak electrolyte |
|---|---|---|
| Λm on dilution | Rises slowly | Rises steeply |
| Λ°m | Found by extrapolation | Found by Kohlrausch's law |
| Example | KCl, NaCl | CH₃COOH, NH₄OH |
Practice questions
- 1Write the SI unit of conductance.
- 2How does the molar conductivity of a strong electrolyte change on dilution?
- 3State Kohlrausch's law.
- 4A cell has G* = 1.29 cm⁻¹ and its sample has R = 250 Ω. Find κ.
- 5Calculate Λm of 0.10 M KCl solution with κ = 0.0129 S cm⁻¹.
- 6Why is an alternating current used to measure conductivity?
- 7Find Λ°m of acetic acid if λ°(H⁺) = 349.6 and λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹.
- 8A 0.05 M solution of a weak acid has Λm = 7.8 and Λ°m = 390.5. Find α and Ka.
Show answers (try them yourself first)
- 1Siemens (S, ohm⁻¹).
- 2It increases slowly, approaching Λ°m.
- 3The limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion.
- 4κ = 1.29 ÷ 250 = 0.00516 S cm⁻¹.
- 5Λm = 0.0129 × 1000 ÷ 0.10 = 129 S cm² mol⁻¹.
- 6D.C. would cause electrolysis and change the solution; A.C. avoids a permanent chemical change.
- 7Λ°m = 349.6 + 40.9 = 390.5 S cm² mol⁻¹.
- 8α = 7.8 ÷ 390.5 = 0.02; Ka = 0.05 × α² ÷ (1 − α) = 2.0 × 10⁻⁵.
Electrolysis and Faraday's laws
Link to Std 11 : In Std 11 you used oxidation numbers; electrolysis forces non-spontaneous redox reactions by supplying electricity.
Spoons are silver-plated by electrolysis, and aluminium metal is made from its ore by passing current through a molten mixture.
In an electrolytic cell an external source drives a non-spontaneous redox reaction; the cathode is negative and the anode is positive. Faraday's first law: the mass deposited is proportional to the charge passed, m = Z I t. In terms of molar mass, m = M I t ÷ (n F) with F = 96500 C mol⁻¹.
Electrolysis and Faraday's laws
In the app you can move this figure with your finger.
Open in the appIn the figure : Positive ions move to the cathode and gain electrons; negative ions move to the anode and lose electrons. Notice the polarity is the reverse of the galvanic cell.
Rule 1: Faraday's laws — Charge Q = I × t (coulombs). Moles of electrons = Q ÷ 96500. Mass = (moles of electrons ÷ n) × M. Second law: for the same charge, the masses of different substances deposited are proportional to their equivalent masses (M ÷ n).
- Easy : A current of 2.0 A for 30 minutes passes Q = 2.0 × 1800 = 3600 C.
- Medium : Moles of electrons = 3600 ÷ 96500 = 0.0373; Cu²⁺ + 2e⁻ → Cu needs 2 electrons per atom, so Cu = 0.01865 mol = 1.184 g.
- Exam-type : To deposit 1 mol of Al from Al³⁺ needs 3 F = 3 × 96500 = 289500 C.
Rule 2: Products of electrolysis — The product depends on the electrolyte and the electrodes. At the cathode the species with the higher E° (easier to reduce) is discharged; at the anode the species easier to oxidise (lower E°) is discharged. Water can also take part.
- Easy : Molten NaCl: sodium at the cathode and chlorine gas at the anode.
- Medium : Aqueous NaCl: H₂ at the cathode (water is easier to reduce than Na⁺) and Cl₂ at the anode.
- Exam-type : Aqueous CuSO₄ with copper electrodes: Cu is deposited at the cathode and the copper anode dissolves, so the blue colour stays the same (this is how copper is refined).
Rule 3: Electroplating and applications — In electroplating the object to be plated is the cathode, the plating metal is the anode, and the electrolyte contains ions of the plating metal. Metals such as Na, Mg, Al are extracted by electrolysis of molten salts.
- Easy : To silver-plate a spoon: spoon = cathode, silver bar = anode, electrolyte = silver nitrate solution.
- Medium : 0.50 A flowing for 1 h deposits mass of Ag = (0.50 × 3600 ÷ 96500) × 108 = 2.015 g.
- Exam-type : Why must a cell be reversed to get Al? Aluminium is below hydrogen in E°, so Al³⁺ cannot be reduced in water; molten cryolite mixture is used.
Moles of electrons = charge ÷ 96500. Match them to the equation for the half-reaction.
Using time in minutes or hours in Q = I × t. Convert time to seconds.
Memory clue: "Q = I t, then ÷ F" — charge in coulombs, then divide by 96500 to get moles of electrons.
Spot the difference
| Point | Galvanic cell | Electrolytic cell |
|---|---|---|
| Reaction | Spontaneous | Non-spontaneous (forced) |
| Anode sign | Negative | Positive |
| Energy | Chemical → electrical | Electrical → chemical |
Practice questions
- 1Write the value of Faraday's constant.
- 2At which electrode does reduction occur in an electrolytic cell?
- 3Name the cathode and anode used in silver plating.
- 4Find the charge passed by 2.0 A for 30 minutes.
- 5How many faradays are needed to deposit 1 mol of Al³⁺ as Al?
- 6What is formed at the anode on electrolysis of aqueous CuSO₄ using inert electrodes?
- 7Find the mass of Cu deposited by 2.0 A in 30 minutes (Cu = 63.5).
- 8Find the mass of silver deposited when 0.50 A flows for 1 hour (Ag = 108).
Show answers (try them yourself first)
- 196500 C mol⁻¹.
- 2Cathode (reduction).
- 3Cathode: the spoon; anode: a silver bar (with AgNO₃ electrolyte).
- 4Q = 2.0 × 1800 = 3600 C.
- 53 F (3 × 96500 C).
- 6Oxygen gas (water is oxidised).
- 7Mass = (3600 ÷ 96500) ÷ 2 × 63.5 = 1.184 g.
- 8Mass = (0.50 × 3600 ÷ 96500) × 108 = 2.015 g.
Batteries and fuel cells
Link to Std 11 : You studied redox reactions and energy changes; batteries are packaged galvanic cells that store this energy for use.
A mobile phone battery can be recharged, while a wall-clock cell is thrown away when it is finished.
A battery is one or more galvanic cells in series. Primary batteries (dry cell, mercury cell) cannot be recharged. Secondary batteries (lead storage, nickel–cadmium) can be recharged by passing current in the reverse direction. A fuel cell produces electricity directly from fuel and oxygen supplied continuously.
In the figure : Hydrogen gives up electrons at the anode, the electrons go through the wire, and oxygen takes them at the cathode; the ions through the electrolyte complete the circuit. Only water forms.
Rule 1: Primary batteries — Dry cell: zinc container is the anode, a graphite rod in MnO₂ and carbon is the cathode, and NH₄Cl–ZnCl₂ paste is the electrolyte; potential is about 1.5 V and it is not rechargeable. Mercury cell: zinc–mercury amalgam anode and HgO cathode with KOH; gives a constant 1.35 V, used in watches and hearing aids.
- Easy : A torch cell cannot be recharged because the products block the electrodes or the reaction is not reversible.
- Medium : Mercury cell: Zn(Hg) + HgO → ZnO + Hg, potential stays at 1.35 V throughout.
- Exam-type : In a dry cell the anode reaction is Zn → Zn²⁺ + 2e⁻ and ammonia from NH₄⁺ forms a complex with Zn²⁺.
Rule 2: Secondary battery: lead storage — Anode: lead; cathode: lead dioxide on lead; electrolyte: about 38 % H₂SO₄. Discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Charging reverses it. One cell gives about 2 V; six cells in series give 12 V.
- Easy : A car battery marked 12 V has 6 cells, each about 2 V: 6 × 2.0 = 12 V.
- Medium : On discharge, sulphuric acid is used up, so its density falls; this is why density is checked.
- Exam-type : Both electrodes are converted to PbSO₄ when the battery is fully discharged; recharging converts them back.
Rule 3: Fuel cells — In an H₂–O₂ fuel cell: anode 2H₂ + 4OH⁻ → 4H₂O + 4e⁻; cathode O₂ + 2H₂O + 4e⁻ → 4OH⁻. Overall 2H₂ + O₂ → 2H₂O. Efficiency is about 70 %, much higher than a thermal power plant, and the only product is water.
- Easy : Fuel cells have been used in space vehicles, where the water formed is used by the crew.
- Medium : ΔrG° for 2H₂ + O₂ → 2H₂O with E° = 1.23 V and n = 4: −4 × 96500 × 1.23 = -474.8 kJ.
- Exam-type : Fuel cells do not need recharging as long as fuel and oxygen are supplied; batteries store a fixed amount of reactants.
Primary: use once. Secondary: rechargeable. Fuel cell: continuous supply of reactants.
Thinking that a fuel cell stores energy. It converts fuel supplied from outside; it does not store it.
Memory clue: "Pb–Pb–Pb, two volts a tub" — a lead cell has two volts; six cells make 12 V.
Spot the difference
| Point | Primary battery | Secondary battery |
|---|---|---|
| Recharge | No | Yes |
| Example | Dry cell, mercury cell | Lead storage, Ni–Cd |
| Reaction | Not reversible in practice | Reversible |
Practice questions
- 1Name the anode of a dry cell.
- 2How many cells does a 12 V lead storage battery have?
- 3What is the only product of the H₂–O₂ fuel cell?
- 4Write the overall reaction when a lead storage battery discharges.
- 5Why does the density of the acid in a lead battery fall on discharge?
- 6Name the electrolyte of a mercury cell and give one use.
- 7Calculate ΔrG° for 2H₂ + O₂ → 2H₂O at E° = 1.23 V.
- 8Give two advantages of a fuel cell over a thermal power plant.
Show answers (try them yourself first)
- 1The zinc container.
- 2Six cells of about 2 V each.
- 3Water.
- 4Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O.
- 5Sulphuric acid is used up in forming PbSO₄ and water is produced, so the solution becomes more dilute.
- 6KOH (with a zinc–mercury amalgam and HgO); used in watches and hearing aids.
- 7ΔrG° = −4 × 96500 × 1.23 = -474.8 kJ.
- 8It is more efficient (about 70 %) and it does not pollute because only water is formed.
Corrosion and its prevention
Link to Std 11 : In Std 11 you learnt oxidation and reduction; corrosion is a slow redox reaction in which a metal is oxidised by air and water.
A bicycle left in the rain gets a reddish-brown coating of rust, which flakes off and weakens the metal.
Corrosion is the slow eating away of a metal by reaction with air and moisture; rusting of iron is the best-known example. It is an electrochemical process: one spot acts as the anode and another as the cathode, with the water drop as the electrolyte.
In the figure : At the anode iron dissolves as Fe²⁺. At the cathode oxygen dissolved in the water is reduced. Fe²⁺ is oxidised further to Fe³⁺ and appears as hydrated iron(III) oxide, the reddish-brown rust.
Rule 1: Mechanism of rusting — Anode: Fe(s) → Fe²⁺(aq) + 2e⁻ (E° = −0.44 V). Cathode: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) (acidity from dissolved CO₂). Overall: 2Fe + O₂ + 4H⁺ → 2Fe²⁺ + 2H₂O. Fe²⁺ → Fe³⁺ in air, giving Fe₂O₃·xH₂O.
- Easy : Rusting needs both oxygen and water; iron kept in dry air or in boiled water without air does not rust.
- Medium : Impure iron rusts faster than pure iron because impurities set up small galvanic cells.
- Exam-type : Why does rusting go faster near the sea? Salt water is a better electrolyte and increases the current of the cell.
Rule 2: Preventing corrosion — Cover the surface with paint, oil or a metal coating; make the iron a cathode in contact with a more active metal (sacrificial protection) or use an alloy.
- Easy : Galvanised iron has a coat of zinc. Zinc (−0.76 V) is more reactive than iron (−0.44 V).
- Medium : Even when scratched, zinc corrodes first and protects the iron; tin plate fails once scratched because iron (−0.44 V) is more active than tin (−0.14 V).
- Exam-type : Magnesium blocks attached to underground pipes or ship hulls are sacrificial anodes that corrode instead of the iron.
Rule 3: Why electrochemistry matters — Electrochemical ideas explain batteries, fuel cells and corrosion, and they are central to the hydrogen economy, in which hydrogen made by electrolysis using renewable electricity serves as a clean fuel.
- Easy : Fuel cells burn hydrogen without giving out carbon dioxide.
- Medium : Electrolysis of water splits H₂O into H₂ and O₂, so hydrogen can be stored as a fuel.
- Exam-type : Compare sacrificial protection and tin plating: a more active coating metal protects iron even when the coating is broken; a less active one does not.
Iron rusts only when oxygen and water are both present, and the coating metal should be more active than iron for sacrificial protection.
Saying that rust is Fe₂O₃ only; it is hydrated: Fe₂O₃·xH₂O.
Memory clue: "Zinc zaps rust" — zinc sits below iron in E°, so zinc corrodes first and saves the iron.
Spot the difference
| Point | Zinc coating (galvanising) | Tin coating |
|---|---|---|
| Metal E° | −0.76 V (more active than Fe) | −0.14 V (less active than Fe) |
| When scratched | Zinc corrodes, iron safe | Iron corrodes faster |
| Type | Sacrificial protection | Barrier only |
Practice questions
- 1Write the formula of rust.
- 2Name two things needed for the rusting of iron.
- 3What is a sacrificial anode?
- 4Write the anode and cathode reactions in rusting.
- 5Why is galvanised iron safe even when scratched?
- 6Why does a tin-plated iron can rust fast when the tin layer is broken?
- 7Explain why pipes buried in soil are connected to magnesium blocks.
- 8Explain why impure iron rusts faster than pure iron.
Show answers (try them yourself first)
- 1Fe₂O₃·xH₂O.
- 2Oxygen (air) and water (moisture).
- 3A more active metal connected to iron so that the more active metal is oxidised instead of iron.
- 4Anode: Fe → Fe²⁺ + 2e⁻; cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O.
- 5Zinc is more active than iron, so zinc corrodes first and the iron acts as the cathode.
- 6Iron (−0.44 V) is more active than tin (−0.14 V), so once exposed, iron becomes the anode and corrodes faster.
- 7Magnesium (E° = −2.36 V) is more active than iron, so it is oxidised instead of the pipe and protects it.
- 8Impurities form tiny electrodes of different potential in the iron, so many small galvanic cells operate and the corrosion current is larger.