Degradation of Materials: Corrosion and Polymer Degradation

This chapter answers Section XE2.6, “Degradation of Materials”: “Electrochemical basis of corrosion of metals: standard electrode potential, galvanic series, Nernst equation, polarization and passivation; forms of corrosion; corrosion prevention. Polymer degradation: swelling and dissolution; bond rupture: radiation, chemical and thermal effects; weathering.” It is Section XE2.6 of XE2 Materials Science, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. It teaches the electrochemical cell behind corrosion, the rate from current density by Faraday’s law, the eight forms of corrosion, the ways of preventing it, and the mechanisms by which polymers swell, dissolve and break down. The numericals are the Nernst potential, the corrosion penetration rate, the mass of a sacrificial anode and the energy of an ultraviolet photon.

1. Electrochemical basis: electrode potential, Nernst equation, galvanic series

Aqueous corrosion needs a complete cell: an anode where the metal is oxidised, M → Mⁿ⁺ + ne⁻, a cathode where electrons are consumed (in aerated neutral water O₂ + 2H₂O + 4e⁻ → 4OH⁻; in acid 2H⁺ + 2e⁻ → H₂), a metallic path for electrons and an electrolyte for ions. The metal dissolves at the anode only. The standard electrode potential E° (volts, against the standard hydrogen electrode at 0 V, 25 °C, unit activity) measures the tendency to be reduced, so a more negative E° means a more active metal.

Selected standard electrode potentials (25 °C)
Half-reactionE° (V)Half-reactionE° (V)
Au³⁺ + 3e⁻ → Au+1.50Fe²⁺ + 2e⁻ → Fe−0.44
Ag⁺ + e⁻ → Ag+0.80Zn²⁺ + 2e⁻ → Zn−0.76
Cu²⁺ + 2e⁻ → Cu+0.34Al³⁺ + 3e⁻ → Al−1.66
2H⁺ + 2e⁻ → H₂0.00Mg²⁺ + 2e⁻ → Mg−2.37

A cell has E_cell = E_cathode − E_anode, and the corroding member is the one with the lower potential. The Nernst equation corrects for concentration: E = E° − (RT/nF) ln Q, i.e. at 25 °C E = E° − (0.0592/n) log₁₀ Q, where Q is the reaction quotient (for M²⁺ + 2e⁻ → M, Q = 1/[M²⁺]). Lowering [Mⁿ⁺] makes the potential more negative, so the metal becomes more active. The galvanic series ranks actual metals and alloys by their measured potentials in a stated environment, usually seawater: unlike the emf series it covers alloys and passive states (stainless steel and titanium sit at the noble end when passive), so it is the practical guide to which metal of a couple corrodes.

ℹ️ Worked: Nernst cell voltage
A Zn|Zn²⁺(0.01 M) || Cu²⁺(1.0 M)|Cu cell at 25 °C, with 2.303RT/F = 0.0592 V. E°_cell = 0.34 − (−0.76) = 1.10 V. Zn + Cu²⁺ → Zn²⁺ + Cu has Q = [Zn²⁺]/[Cu²⁺] = 0.01 and n = 2, so E = 1.10 − (0.0592/2) log₁₀(0.01) = 1.10 + 0.0592 = 1.16 V.

2. Polarization, passivation and the corrosion rate

A corroding metal is a short-circuited cell: the anodic and cathodic reactions run at the same rate on one surface, at the corrosion potential E_corr, with current density i_corr. Polarization is the departure of an electrode potential from its equilibrium value when current flows, by an overpotential η. Activation polarization (slow charge-transfer step) follows the Tafel relation η = a + b log i. Concentration polarization appears when a reactant cannot arrive fast enough: the cathodic current saturates at a limiting current density, set by diffusion of dissolved oxygen. In neutral aerated water, iron corrosion is usually cathodically controlled by this oxygen supply, so more dissolved oxygen or more stirring raises the rate.

Passivation: some metals (Cr, Al, Ti, stainless steels) form a thin, adherent, self-healing oxide film that stops anodic dissolution. On an anodic polarization curve the current rises in the active region, falls sharply at the passivation potential to a very small passive current, and rises again in the transpassive region when the film breaks down. Stainless steels contain more than about 12 wt% Cr for this reason. Passive films fail locally in chloride, which starts pitting.

Faraday’s law converts current density to metal loss: the mass dissolved per unit area per unit time is w = iM/(nF), with M the molar mass, n the electrons per atom and F = 96 485 C/mol. The penetration rate is w/ρ.

ℹ️ Worked: corrosion rate of iron
Iron (M = 55.85 g/mol, n = 2, ρ = 7.87 g/cm³) corrodes at i = 50 μA/cm². w = (50 × 10⁻⁶ × 55.85)/(2 × 96 485) = 1.447 × 10⁻⁸ g cm⁻² s⁻¹. Penetration = w/ρ = 1.839 × 10⁻⁹ cm/s. In one year (3.1536 × 10⁷ s) this is 0.0580 cm = 0.58 mm/yr.
⚠️ Units and valence
Convert μA/cm² to A/cm² and seconds to years before comparing with mm/yr. Use the valence of the ion actually formed (Fe²⁺ gives n = 2); a mistake in n changes the rate in proportion.

3. Forms of corrosion

The eight forms
FormMechanism and feature
UniformAnode and cathode sites shift over the whole surface; most predictable and measured as a penetration rate
GalvanicTwo dissimilar metals in contact in an electrolyte: the more active one corrodes faster; a small anode against a large cathode is the worst area ratio
PittingLocalised, autocatalytic attack, often by chloride on a passive film; acid, chloride-rich solution builds up in the pit
CreviceIn a narrow gap oxygen is used up, giving differential aeration: the oxygen-poor crevice becomes the anode
IntergranularAttack along grain boundaries; sensitised austenitic stainless steel (about 500 to 800 °C) precipitates Cr carbide at boundaries and leaves Cr-depleted strips
Stress corrosion crackingCracking needs a tensile stress and a specific environment (brass in ammonia, austenitic stainless steel in chloride); removing either stops it
Erosion–corrosionFast flow or impingement removes the protective film, so attack accelerates (pipe bends, impellers)
Selective leachingOne element of an alloy is removed, as zinc from brass (dezincification) or iron from grey cast iron (graphitic corrosion), leaving a weak porous residue
⚠️ Why crevice and pitting corrode the inside
In crevice corrosion the anode is the oxygen-starved interior and the exposed outer surface is the cathode, the opposite of what intuition suggests. Sensitisation is prevented by low-carbon grades, by stabilising the steel with Ti or Nb, or by solution annealing.

4. Corrosion prevention

Coatings separate metal from environment. A barrier coating (paint, polymer, enamel, tin plate) only works while it is unbroken. Sacrificial coating: galvanised zinc is more active than steel (−0.76 V against −0.44 V), so at a scratch zinc corrodes and protects the steel. Tin is nobler than iron in aerated neutral water, so at a break in tin plate the steel corrodes faster than if it were bare. Cathodic protection makes the whole structure the cathode: a sacrificial anode (Mg, Zn or Al) is electrically connected to it and dissolves in its place, while an impressed current system drives current from an external DC source through a nearly inert anode (graphite, for example). Sacrificial anodes need no power and suit small or well-coated structures; impressed current suits large structures and high-resistivity soils.

Inhibitors are small additions to the environment that slow corrosion: anodic inhibitors strengthen the passive film, cathodic inhibitors slow the cathodic reaction, and many organic inhibitors adsorb on the surface. Design avoids crevices and trapped water, allows drainage, avoids a small anode against a large cathode, insulates dissimilar metals, and keeps tensile stresses and flow velocities low. Choosing a more resistant alloy is also a design choice.

ℹ️ Worked: anode consumption
A zinc anode (M = 65.4 g/mol, n = 2) supplies 0.5 A continuously for one year (3.1536 × 10⁷ s). Charge = 1.577 × 10⁷ C; moles of Zn = 1.577 × 10⁷/(2 × 96 485) = 81.7 mol; mass = 81.7 × 65.4 = 5.34 × 10³ g = 5.34 kg.

5. Polymer degradation: swelling and dissolution

Polymers do not corrode electrochemically; they degrade by interaction with liquids and by chain scission. Swelling: small solvent molecules diffuse between the chains, push them apart and enlarge the volume, plasticising the polymer (softer, weaker, lower T_g). It is favoured when the solvent and polymer have similar solubility parameters (“like dissolves like”) and is limited in a cross-linked polymer, which swells to an equilibrium amount but cannot dissolve because the network holds it together. A linear, uncross-linked polymer can go on to dissolution as chains disentangle and diffuse into the liquid: it swells first, then dissolves. Dissolution is slower for higher molecular weight and for higher crystallinity, and crystalline polymers such as polyethylene resist solvents at room temperature.

⚠️ Swelling is not dissolution
A cross-linked rubber placed in a good solvent swells strongly and never dissolves. Swelling raises volume and mass, but the chains are not broken; the effect is largely physical and reversible on drying.

6. Bond rupture: radiation, chemical and thermal effects; weathering

Radiation: a photon can break a covalent bond if its energy exceeds the bond energy. The photon energy per mole is E = N_A hc/λ; at 300 nm it is 399 kJ/mol, more than the roughly 350 kJ/mol of a C–C bond, so ultraviolet in sunlight breaks bonds while visible light (about 300 kJ/mol at 400 nm) mostly does not. The broken bonds give free radicals that cause chain scission (lower molecular weight, loss of strength, brittleness) or cross-linking (stiffening, embrittlement); the balance depends on the polymer.

Chemical effects: oxygen and ozone attack by free-radical oxidation, which antioxidants slow; ozone cracks unsaturated rubbers (C=C in the backbone) that are under tensile strain; hydrolysis breaks the backbone of polyesters and polyamides in hot, moist conditions; strong acids and bases attack susceptible links. Thermal effects: at high temperature chains break by random scission, as in polyethylene; PMMA depolymerises (“unzips”) back to monomer; PVC eliminates HCl from its side groups, discolouring and weakening it, which is why PVC needs heat stabilisers.

Weathering is the combined action of ultraviolet light, oxygen, moisture, temperature cycles and pollutants on a polymer outdoors. It shows as colour change, surface chalking and cracking, and loss of impact strength, starting at the exposed surface. It is slowed by UV absorbers, antioxidants and pigments such as carbon black that screen the light.

ℹ️ Worked: photon energy
Take h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s and N_A = 6.022 × 10²³ mol⁻¹. At λ = 300 nm, E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸ × 6.022 × 10²³)/(300 × 10⁻⁹) = 3.99 × 10⁵ J/mol = 399 kJ/mol. A photon of 400 nm carries 299 kJ/mol.

Key takeaways

  • Corrosion needs an anode, a cathode, an electron path and an electrolyte; E_cell = E_cathode − E_anode, and the more negative metal corrodes. The galvanic series, not the emf series, ranks real alloys in seawater.
  • Nernst: E = E° − (0.0592/n) log₁₀ Q at 25 °C; lowering the ion concentration makes a metal electrode more negative.
  • Rate from Faraday’s law: w = iM/(nF) and penetration = w/ρ; oxygen diffusion usually limits corrosion in neutral aerated water, and passive films (Cr, Al, Ti) stop anodic dissolution.
  • Learn each form by its signature: small anode area in galvanic, differential aeration in crevice, Cr depletion in intergranular, tensile stress plus a specific environment in SCC, dezincification in selective leaching.
  • Prevention: coatings (zinc is sacrificial, tin is not), cathodic protection by sacrificial anode or impressed current, inhibitors, and design that avoids crevices and bad area ratios.
  • Polymers swell and dissolve by solvent uptake (cross-linked ones only swell) and degrade by UV, ozone, hydrolysis and heat through chain scission or cross-linking; weathering is the combined outdoor effect.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Zinc (E° = −0.76 V) and copper (E° = +0.34 V) are electrically connected and immersed in the same aerated aqueous electrolyte. Which statement is correct?

    1. Copper is the anode and dissolves, because it has the higher potential
    2. Neither metal corrodes, since both are metals and conduct electrons
    3. Zinc is the anode and corrodes, while copper is the cathode
    4. Both corrode at equal rates because the same current flows through each
    Show answer

    Answer: C — Zinc is the anode and corrodes, while copper is the cathode

    The metal with the more negative potential is the anode and is oxidised; zinc dissolves and the electrons flow to copper where oxygen is reduced. The statement that copper is the anode reverses the rule. The same current does flow through both, but it is the anodic current that dissolves metal, and that is entirely on zinc.

  2. A copper rod dips in a CuSO₄ solution. According to the Nernst equation, if the Cu²⁺ concentration is lowered from 1 M to 0.001 M at 25 °C, the electrode potential

    1. rises by about 0.09 V, so copper becomes nobler
    2. falls by about 0.09 V, so copper becomes more active
    3. falls by about 0.18 V because three decades of concentration are involved
    4. does not change, since E° alone fixes the potential of a metal
    Show answer

    Answer: B — falls by about 0.09 V, so copper becomes more active

    E = E° + (0.0592/2) log₁₀[Cu²⁺]. Going from 1 M to 0.001 M gives 0.0296 × (−3) = −0.089 V, a fall of about 0.09 V, so copper becomes more active. The potential rises when the concentration rises, not when it is lowered. The 0.18 V figure forgets to divide by n = 2, and E° alone applies only at unit activity.

  3. Stainless steel resists corrosion in many environments mainly because

    1. chromium raises its standard electrode potential far above that of iron, making the alloy a noble metal
    2. it forms a thin, self-healing chromium-rich oxide film that stops anodic dissolution
    3. the carbon in the steel acts as a sacrificial anode that protects the iron matrix from attack
    4. it contains no iron at all, so no anodic reaction can occur anywhere on its surface
    Show answer

    Answer: B — it forms a thin, self-healing chromium-rich oxide film that stops anodic dissolution

    Chromium (above about 12 wt%) forms a protective passive film that re-forms if scratched, which drops the anodic current to a very small passive value. Chromium is itself an active metal with a negative E°, so a positive shift of potential in the emf series is not the cause. Carbon in steel does not protect iron, and stainless steel is mostly iron.

  4. Mild steel corrodes in neutral water, and the cathodic reaction O₂ + 2H₂O + 4e⁻ → 4OH⁻ is limited by the diffusion of dissolved oxygen to the surface. If the water is stirred more vigorously, the corrosion current is expected to

    1. remain unchanged, because the anodic reaction alone fixes the corrosion rate of the steel
    2. fall, because stirring sweeps the dissolved metal ions away from the metal surface
    3. fall to zero, since the stirring sweeps all of the oxygen away from the metal
    4. rise, because the limiting current of oxygen reduction increases
    Show answer

    Answer: D — rise, because the limiting current of oxygen reduction increases

    Stirring thins the diffusion layer, raising the limiting oxygen current. When the cathodic reaction controls the rate, a larger cathodic limiting current raises the corrosion current by the same amount. The anodic reaction cannot fix the rate if the cathode is the limiting step, and stirring brings oxygen towards the surface rather than away from it.

  5. A galvanic cell Zn | Zn²⁺ (0.01 M) || Cu²⁺ (1.0 M) | Cu operates at 25 °C. Take E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V and 2.303RT/F = 0.0592 V. The cell potential is ____ V (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 1.16

    E°_cell = 0.34 − (−0.76) = 1.10 V. For Zn + Cu²⁺ → Zn²⁺ + Cu, Q = 0.01/1.0 and n = 2, so E = 1.10 − (0.0592/2) log₁₀(0.01) = 1.10 + 0.0592 = 1.159, i.e. 1.16 V. Inverting Q would give 1.04 V.

  6. Iron corrodes uniformly, dissolving as Fe²⁺, at a current density of 50 μA/cm². Take M = 55.85 g/mol, density 7.87 g/cm³, F = 96 485 C/mol and 1 year = 3.1536 × 10⁷ s. The penetration (thickness loss) rate is ____ mm per year (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.58

    Mass loss per area and time: w = iM/(nF) = (5 × 10⁻⁵ × 55.85)/(2 × 96 485) = 1.447 × 10⁻⁸ g cm⁻² s⁻¹. Thickness rate = w/ρ = 1.839 × 10⁻⁹ cm/s; times 3.1536 × 10⁷ s gives 0.0580 cm = 0.58 mm/yr. Using n = 3 would give 0.39.

  7. Which of the following statements about galvanic corrosion and electrode potentials are correct?

    1. The galvanic series is measured in a stated environment, such as seawater, and includes alloys and passive metals
    2. A small anode area connected to a large cathode area gives a high anodic current density and rapid attack
    3. E_cell = E_anode − E_cathode, so a spontaneous cell has a negative potential
    4. The standard hydrogen electrode is the reference taken as 0 V in the emf series
    Show answer

    Answer: A — The galvanic series is measured in a stated environment, such as seawater, and includes alloys and passive metals; B — A small anode area connected to a large cathode area gives a high anodic current density and rapid attack; D — The standard hydrogen electrode is the reference taken as 0 V in the emf series

    The galvanic series is environment-specific and includes alloys, the area ratio sets the current density on the anode, and the hydrogen electrode defines 0 V. The relation for the cell potential is reversed: E_cell = E_cathode − E_anode, which is positive for a spontaneous cell.

  8. A welded austenitic stainless steel with 0.08 wt% C shows attack along the grain boundaries next to the weld, in the zone held at about 500 to 800 °C. The most likely cause is

    1. chromium carbide precipitating at the boundaries, which leaves strips depleted in chromium
    2. differential aeration under the weld spatter, which makes the oxygen-poor interior the anode
    3. selective removal of nickel from the alloy to leave a weak porous residue behind
    4. a small cathode area in contact with a large anode area in the weld metal
    Show answer

    Answer: A — chromium carbide precipitating at the boundaries, which leaves strips depleted in chromium

    This is sensitisation: Cr₂₃C₆-type carbide forms at the boundaries and the neighbouring zone falls below the Cr level needed for passivity, so the boundaries corrode. Crevice attack is a different form with a different location. No selective leaching of nickel occurs here, and a small cathode on a large anode would reduce, not raise, localised attack.

  9. In crevice corrosion of a stainless steel plate bolted to another surface in aerated seawater, the region that acts as the anode is

    1. the exposed outer surface, where oxygen is plentiful
    2. the oxygen-depleted metal inside the narrow gap
    3. the bolt, because fasteners are always anodic to plates
    4. the whole plate equally, since the potential is uniform
    Show answer

    Answer: B — the oxygen-depleted metal inside the narrow gap

    Oxygen consumed inside the gap is not replaced, so the interior is oxygen-poor and becomes the anode, with the exposed surface as the cathode. The outer surface, being oxygen-rich, is the cathode. Whether a fastener is anodic depends on the metals, not on its being a fastener, and a uniform potential would give no localised attack.

  10. A brass component fails by cracking in a humid ammonia-containing atmosphere although its design load is well below yield. Which change is most certain to prevent the failure?

    1. Painting the part, since the cracks start from general surface rusting and coating stops it
    2. Stress-relief annealing, or removing the ammonia from the environment
    3. Making the cross-section larger so that the applied stress falls below the endurance limit
    4. Connecting the part to a more noble metal so as to make the brass the cathode
    Show answer

    Answer: B — Stress-relief annealing, or removing the ammonia from the environment

    Stress corrosion cracking needs both a tensile stress (applied or residual) and a specific environment, here brass with ammonia. Removing either one stops it, and a stress-relief anneal removes the residual stress. A larger section lowers only the applied stress; residual stress remains and the crack can still grow at low stress. Paint can be breached, and coupling to a noble metal would make brass the anode, not the cathode.

  11. Which of the following statements about corrosion prevention are correct?

    1. A sacrificial anode for steel must have a more negative potential than steel, as magnesium or zinc has
    2. An impressed current system needs an external DC source and an anode that is nearly inert, such as graphite
    3. Cathodic protection works by making the structure the anode of a cell
    4. Anodic inhibitors act by strengthening the passive film on the metal
    Show answer

    Answer: A — A sacrificial anode for steel must have a more negative potential than steel, as magnesium or zinc has; B — An impressed current system needs an external DC source and an anode that is nearly inert, such as graphite; D — Anodic inhibitors act by strengthening the passive film on the metal

    The anode must be more active than steel, an impressed current system needs a DC supply and an inert anode, and anodic inhibitors promote passivity. Cathodic protection does the opposite of the claim that it makes the structure the anode: it makes the structure the cathode, with the sacrificial or inert anode elsewhere.

  12. A zinc sacrificial anode (M = 65.4 g/mol, dissolving as Zn²⁺) supplies a steady current of 0.5 A to a pipeline for one year of 365 days (3.1536 × 10⁷ s). Take F = 96 485 C/mol. Assuming all the charge comes from zinc dissolution, the mass of zinc consumed is ____ kg (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 5.34

    Charge = 0.5 × 3.1536 × 10⁷ = 1.5768 × 10⁷ C. Moles Zn = Q/(nF) = 1.5768 × 10⁷/(2 × 96 485) = 81.71 mol. Mass = 81.71 × 65.4 = 5344 g = 5.34 kg. Forgetting n = 2 would give 10.69 kg.

  13. A steel sheet is coated with zinc in one case and with tin in the other. In aerated neutral water, a scratch exposes the steel in each. Which statement is correct?

    1. Zinc protects the steel by corroding in its place; at a tin scratch the steel corrodes faster than bare steel
    2. Both coatings protect the exposed steel equally well, because each covers most of the whole surface
    3. Tin protects the exposed steel because tin is nobler than iron, while zinc speeds up attack on steel
    4. Neither coating affects the exposed steel, since only the bare metal itself determines its corrosion rate in the water
    Show answer

    Answer: A — Zinc protects the steel by corroding in its place; at a tin scratch the steel corrodes faster than bare steel

    Zinc (−0.76 V) is more active than iron (−0.44 V), so it becomes the anode and cathodically protects the scratch. Tin (−0.14 V) is nobler than iron, so steel at a break becomes a small anode against a large tin cathode and corrodes quickly. Equal protection ignores the potentials, and the claim that tin protects reverses which metal is the anode.

  14. A vulcanised (cross-linked) rubber block and a sample of the same polymer without cross-links are both placed in a good solvent for the polymer. Which description is correct?

    1. Both swell to a finite equilibrium amount and neither of them dissolves in the liquid
    2. The cross-linked rubber dissolves faster than the linear polymer because its chains are shorter
    3. The cross-linked rubber swells to a limit, while the linear polymer swells and then dissolves
    4. Neither swells, because even a good solvent cannot enter a dense solid polymer
    Show answer

    Answer: C — The cross-linked rubber swells to a limit, while the linear polymer swells and then dissolves

    Solvent diffuses into both and swells them. In the network the cross-links hold the chains together, limiting swelling to an equilibrium value; the linear chains can disentangle and pass into solution, so that polymer dissolves. Cross-links make chains effectively longer, not shorter, and the solvent does enter, which is why swelling occurs.

  15. Which of the following statements about polymer degradation are correct?

    1. PMMA can depolymerise on heating, giving back its monomer
    2. Unsaturated rubbers under tensile strain are cracked by ozone attacking the C=C bonds
    3. Polyesters and polyamides are unaffected by hot, moist conditions because their backbones are carbon–carbon bonds only
    4. Ultraviolet photons can break covalent bonds because their energy per mole exceeds a typical C–C bond energy
    Show answer

    Answer: A — PMMA can depolymerise on heating, giving back its monomer; B — Unsaturated rubbers under tensile strain are cracked by ozone attacking the C=C bonds; D — Ultraviolet photons can break covalent bonds because their energy per mole exceeds a typical C–C bond energy

    PMMA unzips on heating, ozone cracks strained unsaturated rubber, and UV photons of 300 nm carry about 399 kJ/mol, above the roughly 350 kJ/mol of a C–C bond. The statement about polyesters and polyamides is wrong: their ester and amide links are hydrolysed in hot, moist conditions, and their backbones are not carbon–carbon only.

  16. An ultraviolet photon of wavelength 300 nm is absorbed by a polymer. Take h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s and N_A = 6.022 × 10²³ mol⁻¹. The energy of one mole of such photons is ____ kJ/mol (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 399

    E = N_A hc/λ = 6.022 × 10²³ × 6.626 × 10⁻³⁴ × 3.00 × 10⁸/(300 × 10⁻⁹) = 3.99 × 10⁵ J/mol = 399 kJ/mol. Forgetting N_A gives only 6.6 × 10⁻¹⁹ J per photon; using 300 as metres would be off by a factor of 10⁹.

  17. An outdoor polyethylene pipe becomes chalky and brittle after years in sunlight, and the damage begins at the exposed surface. Which additive most directly slows this weathering?

    1. A plasticiser, which lowers T_g and so makes the chains more mobile and easier to move
    2. Carbon black with a UV absorber, which screens out the ultraviolet light
    3. A filler of chalk powder, which supplies calcium ions that react with the moisture
    4. A solvent, which swells the chains and lets the trapped oxygen leave the pipe
    Show answer

    Answer: B — Carbon black with a UV absorber, which screens out the ultraviolet light

    Weathering is driven mainly by UV light with oxygen and moisture; carbon black and UV absorbers screen or absorb the light before it breaks bonds, and antioxidants stop the radical chain. A plasticiser increases mobility and does nothing to prevent photo-oxidation. Chalk does not act as a UV screen in that way, and a solvent would swell and weaken the pipe.