Exercises — Corrosion — electrochemical mechanism; cathodic protection, galvanization

What we're deciding: which region loses electrons and which gains them.

• Where iron dissolves, iron atoms are turning into ${\text{Fe}}^{2+}$ and releasing electrons. Losing electrons is oxidation, and the oxidation site is the anode. $\text{P (anode)}:\text{Fe}(s)\to {\text{Fe}}^{2+}(aq)+2e^{-}$
• Where oxygen is consumed, ${\text{O}}_2$ is grabbing those electrons. Gaining electrons is reduction, and the reduction site is the cathode. $\text{Q (cathode)}:{\text{O}}_2(g)+2{\text{H}}_2\text{O}(l)+4e^{-}\to 4{\text{OH}}^{-}(aq)$ Answer: P = anode, Q = cathode.

Corrosion is oxidation of the metal. A metal is oxidised easily when it is reluctant to stay reduced — i.e. when its reduction potential is negative.

• Iron: $E°=-0.44\text{ V}$ (negative) → gives up electrons willingly → corrodes readily.
• Gold: $E°=+1.50\text{ V}$ (strongly positive) → clings to its electrons → essentially does not corrode. Answer: Iron corrodes; gold does not.

Why this formula: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$ tells us the "push" driving electrons around the loop; positive push = spontaneous.

• Cathode (reduction, oxygen): $E°=+0.40\text{ V}$
• Anode (oxidation, iron): its reduction value is $-0.44\text{ V}$ $E{°}_{\text{cell}}=0.40-(-0.44)=+0.84\text{ V}$ Positive → the process releases energy → spontaneous. (Confirm with $\Delta G°=-nFE°<0$.) Answer: $E{°}_{\text{cell}}=+0.84\text{ V}$, spontaneous.

Why $\Delta G°$: it converts the cell voltage into an energy per mole of reaction — the language of Gibs Free Energy and thermodynamic spontaneity. $\Delta G°=-nFE°=-(4)(96500)(0.84)\text{ J}$ $=-324,240\text{ J}\approx -324\text{ kJ}$ Negative → spontaneous, consistent with 2.1. Answer: $\Delta G°\approx -324\text{ kJ}$.

Why Nernst: standard $E°$ assumes 1 M / 1 atm everything. Real seawater is dilute and basic, so we correct with Nernst Equation: $E=E°-\frac{0.059}{n}\log Q,Q=\frac{[{\text{Fe}}^{2+}{]}^2}{[{\text{H}}^{+}{]}^4{P}_{{\text{O}}_2}}$ (Pure solids/liquids like Fe and H₂O don't appear in $Q$.)

Step 1 — reaction quotient: $Q=\frac{(10^{-6}{)}^2}{(10^{-8}{)}^4(0.21)}=\frac{10^{-12}}{10^{-32}\times 0.21}=\frac{10^{20}}{0.21}\approx 4.76\times 10^{20}$ $\log Q\approx 20.68$ Step 2 — correction term: $\frac{0.059}{4}\times 20.68=0.01475\times 20.68\approx 0.305\text{ V}$ Step 3: $E=1.67-0.305\approx +1.37\text{ V}$ Still strongly positive → corrosion persists in seawater. Answer: $E_{\text{cell}}\approx +1.37\text{ V}$.

Key idea (differential aeration): the region starved of oxygen becomes the anode; the oxygen-rich region becomes the cathode.

• Exposed part: plenty of ${\text{O}}_2$ → good cathode → ${\text{O}}_2+2{\text{H}}_2\text{O}+4e^{-}\to 4{\text{OH}}^{-}$ happens here, metal is protected.
• Buried part: little ${\text{O}}_2$ → forced to be the anode → $\text{Fe}\to {\text{Fe}}^{2+}+2e^{-}$ → this metal dissolves. The oxygen-poor buried metal corrodes faster. Answer: the buried, oxygen-poor region is the anode and pits fastest.

Criterion: the sacrificial metal must be more easily oxidised than iron, i.e. it must have $E°<E{°}_{\text{Fe}}=-0.44\text{ V}$. Then it becomes the anode and corrodes instead of the iron, which is forced to be the cathode.

• Cu $+0.34>-0.44$ ✗ (nobler — would make Fe the anode!)
• Sn $-0.14>-0.44$ ✗ (still nobler than Fe)
• Zn $-0.76<-0.44$ ✓
• Mg $-2.37<-0.44$ ✓ (most negative → strongest protector) Answer: Zn and Mg work; Mg protects most strongly.

Anode (Zn, the sacrifice): $\text{Zn}(s)\to {\text{Zn}}^{2+}(aq)+2e^{-}$ Cathode (oxygen reduction, on the Fe surface): ${\text{O}}_2(g)+2{\text{H}}_2\text{O}(l)+4e^{-}\to 4{\text{OH}}^{-}(aq)$ Balance electrons (×2 on Zn), add: $2\text{Zn}+{\text{O}}_2+2{\text{H}}_2\text{O}\to 2{\text{Zn}}^{2+}+4{\text{OH}}^{-}$ $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=0.40-(-0.76)=+1.16\text{ V}$ Answer: $E{°}_{\text{cell}}=+1.16\text{ V}$; Zn dissolves, Fe is spared.

Compare the coating's $E°$ to iron's $-0.44\text{ V}$ (see figure: the green bar sits below iron, the red bar above).

• Zinc ($-0.76<-0.44$): even at a scratch, Zn is still the anode → Zn corrodes, Fe stays cathode. Protection continues. This is genuine cathodic protection (sacrificial coating).
• Tin ($-0.14>-0.44$): Sn is nobler than Fe. At a scratch, the exposed Fe becomes the anode of a small Fe–Sn cell and corrodes faster than bare iron would. Tin only protects as long as it's an unbroken barrier. Answer: Zinc keeps protecting; scratched tin accelerates iron corrosion.

Why Faraday: Faraday's Laws link charge passed to moles of metal dissolved: $\text{mass}=\frac{MIt}{nF}$. We invert it to solve for $t$.

Step 1 — total charge to consume all Mg: $\text{moles Mg}=\frac{12000\text{ g}}{24\text{ g/mol}}=500\text{ mol}$ $Q=n(\text{mol})F=2\times 500\times 96500=9.65\times 10^7\text{ C}$ Step 2 — time at $0.50\text{ A}$ ($I=Q/t$): $t=\frac{Q}{I}=\frac{9.65\times 10^7}{0.50}=1.93\times 10^8\text{ s}$ Step 3 — convert to years ($1\text{ yr}\approx 3.156\times 10^7\text{ s}$): $t=\frac{1.93\times 10^8}{3.156\times 10^7}\approx 6.1\text{ years}$ Answer: $\approx 6.1$ years.

Why energy: impressed-current protection uses an external DC source (parent note, method B), so it costs electricity, not metal.

Charge in one year: $Q=It=0.50\times 3.156\times 10^7=1.578\times 10^7\text{ C}$. Energy: $W=QV=1.578\times 10^7\times 2.0=3.156\times 10^7\text{ J}=3.156\times 10^4\text{ kJ}$. In kWh: $W=\frac{3.156\times 10^7}{3.6\times 10^6}\approx 8.77\text{ kWh}$. Trade-off: sacrificial = no running power but you replace the block (here every ~6 yr); impressed-current = small ongoing energy bill (~9 kWh/yr) but the inert anode lasts far longer. Choice depends on structure size and access. Answer: $\approx 3.16\times 10^4\text{ kJ}\approx 8.8\text{ kWh}$ per year.

Thermodynamics says Al should corrode even more eagerly than Fe (more negative $E°$). Kinetics overrides it: Al instantly forms a thin, dense, adherent ${\text{Al}}_2{\text{O}}_3$ film that seals the surface — the metal beneath is cut off from oxygen and water. This is passivation (see Passivation). By contrast, iron's oxide (rust) is porous and flaky, offering no seal, so corrosion tunnels inward. Answer: Al is thermodynamically reactive but kinetically protected by a passivating ${\text{Al}}_2{\text{O}}_3$ layer; Fe's oxide fails to passivate.

Sacrificial anode must have $E°$ compared to protected metal ::: more negative (more easily oxidised) Scratched tin coating on iron does what ::: accelerates iron corrosion (Fe becomes anode) Buried (low-O₂) part of a plate acts as ::: the anode (differential aeration) Al resists corrosion despite negative $E°$ because of ::: passivation (dense Al₂O₃ film) $E{°}_{\text{cell}}$ formula ::: $E{°}_{\text{cathode}}-E{°}_{\text{anode}}$ (both as reduction potentials)