Corrosion — electrochemical mechanism; cathodic protection, galvanization: Worked examples

This page drills the parent topic through worked examples. Before we compute anything, we lay out every kind of case corrosion problems come in — so when you meet a new one on an exam, you already know which box it lives in.

Everything here rests on one master formula. Let us earn each symbol.

Definition The three new counting symbols (learn these FIRST)

Before any formula uses them, here is what the little letters mean:

$n$ = the number of electrons the balanced overall reaction moves. Just count them.
$F$ = Faraday's constant $\approx 96500$ coulombs per mole of electrons (see Faraday's Laws) — the "price tag" converting moles of electrons into charge.
$Q$ = the reaction quotient: products' concentrations over reactants', each raised to its coefficient. It answers "how far from standard conditions are we right now?"
$P_{O_{2}}$ = the partial pressure of oxygen gas (in atmospheres). Why a pressure and not a concentration? Because $O_{2}$ is a gas, and a gas's "amount available to react" is measured by how hard it pushes on the surface — its pressure — not by molarity. So for gases we write $P$ , for dissolved ions we write $[]$ .

Definition The cell-potential formula, symbol by symbol

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

$E^{\circ}$ (read "E-standard") is a number in volts that measures how badly a reaction wants to pull electrons IN (get reduced). Bigger = greedier for electrons. It is tabulated in Standard Electrode Potentials.
cathode = the site where reduction (gaining electrons) happens.
anode = the site where oxidation (losing electrons) happens.
We subtract the anode value because the anode reaction runs backwards (oxidation = reverse of the reduction we look up), and reversing a reaction flips the sign of its potential.

A positive $E_{cell}^{\circ}$ means "this happens on its own" — spontaneous. Why? Because $Δ G^{\circ} = - n F E_{cell}^{\circ}$ (from Gibs Free Energy, and now that $n$ and $F$ are defined above, this reads: charge moved × voltage = energy released), so positive $E$ makes $Δ G$ negative, and negative $Δ G$ is nature's green light. Note the subscript "cell": it is the whole cell's potential, not any single electrode's.

Definition The key half-reactions we keep reusing

Every example below is built from these couples. Learn to read each one left-to-right as a reduction (electrons on the left): $Fe^{2 +} (a q) + 2 e^{-} ⇌ Fe (s) E^{\circ} = - 0.44 V$ $O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} ⇌ 4 OH^{-} (a q) E^{\circ} = + 0.40 V (neutral/basic)$ $O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} ⇌ 2 H_{2} O (l) E^{\circ} = + 1.23 V (acidic)$ $2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 OH^{-} (a q) E^{\circ} = - 0.83 V (neutral, basic)$ In acid the hydrogen line is written $2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g)$ , $E^{\circ} = 0.00 V$ . Golden rule of matching: the oxygen cathode has two standard values — $+ 0.40$ V if you write it with $OH^{-}$ , and $+ 1.23$ V if you write it with $H^{+}$ . You must use the one whose species appear in your overall reaction. Mixing them is the classic blunder we fix in Example 3. When a metal corrodes, it runs the Fe line backwards (oxidation, $Fe \to Fe^{2 +} + 2 e^{-}$ ), and one of the other lines forwards (the cathode that swallows those electrons). Which cathode wins depends on what's dissolved — oxygen, or, when oxygen is gone, water/H⁺ itself.

The scenario matrix

Every corrosion / protection problem you will meet is one of these cells:

#	Case class	What makes it tricky	Hit by
A	Standard, spontaneous ( $E_{cell}^{\circ} > 0$ )	plain sign bookkeeping	Ex 1
B	Choosing a protector (sign comparison across a list)	which $E^{\circ}$ is below iron's	Ex 2
C	Non-standard conditions (Nernst, big $Q$ )	powers, logs, negative exponents	Ex 3
D	Degenerate: $E_{cell}^{\circ} \approx 0$ (metal vs itself)	what happens when nothing wins	Ex 4
E	Wrong-way case ( $E_{cell}^{\circ} < 0$ )	the "protected" metal would actually corrode	Ex 5
F	Zero-oxygen limit ( $P_{O_{2}} \to 0$ ): the cathode switches to H₂O/H₂	O₂ route dies, hydrogen route takes over	Ex 6
F′	Acidic corrosion (H⁺ is the cathode, no O₂ needed)	reducing acid conditions, $E^{\circ} = 0.00$ V cathode	Ex 6b
G	Faraday word problem (mass lost over time)	connect current, time, and grams	Ex 7
H	Exam twist (galvanization scratch, area effect)	conceptual trap, no clean number	Ex 8

Now we cover each cell.

Worked example Example 1 — Cell A: standard spontaneous corrosion

Iron in neutral aerated water. Half-cells: $E_{Fe^{2 +} / Fe}^{\circ} = - 0.44 V$ , $E_{O_{2} / OH^{-}}^{\circ} = + 0.40 V$ . Find $E_{cell}^{\circ}$ and say whether iron rusts.

Forecast: Guess first — will $E_{cell}^{\circ}$ be positive or negative? (Iron rusts in the real world, so...)

Write the two half-reactions explicitly. Anode (iron, reversed): $Fe \to Fe^{2 +} + 2 e^{-}$ . Cathode (oxygen, forward): $O_{2} + 2 H_{2} O + 4 e^{-} \to 4 OH^{-}$ . Why this step? Seeing the species tells you exactly which one loses electrons and which gains them — oxygen ( $+ 0.40$ ) is greedier than iron ( $- 0.44$ ), so oxygen is the cathode.
Plug in: $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.40 - (- 0.44) = + 0.84 V$ . Why this step? Two minuses becoming a plus is the whole reason a big driving force appears.
Interpret: positive → spontaneous → iron corrodes.

The figure below shows exactly this: a single drop of water on iron becomes a tiny battery. Look at the red outline — that water film is the electrolyte that lets ions move between the black anode dot (Fe dissolving) and the black cathode dot (O₂ being consumed). Follow the horizontal black arrow: electrons crawl through the metal from anode to cathode.

Verify: $Δ G_{cell}^{\circ} = - n F E_{cell}^{\circ} = - (4) (96500) (0.84) \approx - 3.24 \times 1 0^{5} J < 0$ . Negative $Δ G_{cell}^{\circ}$ confirms spontaneity. ✓

Worked example Example 2 — Cell B: pick the sacrificial anode

An iron tank needs a sacrificial anode. Candidates: Cu ( $+ 0.34$ ), Ni ( $- 0.25$ ), Zn ( $- 0.76$ ), Mg ( $- 2.37$ ), all in volts. Which qualify?

Forecast: A protector must be hungrier to give up electrons than iron. Which of these are "below" iron's $- 0.44$ ?

State the rule: a sacrificial anode needs $E^{\circ} < E_{Fe}^{\circ} = - 0.44 V$ . Why this step? More negative $E^{\circ}$ = wants to be reduced less = wants to oxidize more. That metal corrodes first and spares the iron.
Compare each candidate against the $- 0.44$ threshold:
- Cu $+ 0.34 > - 0.44$ → ✗ (Cu is nobler; iron would corrode to protect Cu — backwards!)
- Ni $- 0.25 > - 0.44$ → ✗ (still above iron)
- Zn $- 0.76 < - 0.44$ → ✓
- Mg $- 2.37 < - 0.44$ → ✓ Why this step? We are literally sorting the list into "below the iron line" (works) and "above the iron line" (useless or harmful) — one number comparison per metal, nothing more.
Answer: Zn and Mg work.

Verify: Cell potential for Mg protecting Fe: $E_{cell}^{\circ} = E_{cathode(Fe)}^{\circ} - E_{anode(Mg)}^{\circ}$ . Using the oxygen cathode $+ 0.40$ : $0.40 - (- 2.37) = + 2.77 V > 0$ , hugely spontaneous → Mg sacrifices itself. ✓ For Cu as anode: $0.40 - 0.34 = + 0.06$ , tiny, and worse, connecting Cu makes iron the anode instead — matching our ✗.

Worked example Example 3 — Cell C: non-standard, big

$Q$ (Nernst) — done the CORRECT way Iron pipe in seawater, $pH = 8$ , so $[H^{+}] = 1 0^{- 8} M$ . Also $[Fe^{2 +}] = 1 0^{- 6} M$ , $P_{O_{2}} = 0.21 atm$ . Find $E_{cell}$ .

Forecast: With so few $H^{+}$ ions (acidic reactant is scarce), will corrosion speed up or slow down versus standard?

Choose ONE consistent set of half-reactions. Because we are given $[H^{+}]$ , we must write the overall reaction with $H^{+}$ in it — so we use the acidic oxygen cathode $O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O$ , $E^{\circ} = + 1.23 V$ , not the $+ 0.40$ V basic one. Overall reaction: $2 Fe + O_{2} + 4 H^{+} \to 2 Fe^{2 +} + 2 H_{2} O .$ Why this step? You may only combine a standard potential with a reaction quotient written in the same species. Since $Q$ will contain $[H^{+}]$ , the cathode potential must be the $H^{+}$ -based one. Mixing $+ 0.40$ V with an $H^{+}$ quotient (as sloppy notes do) is simply wrong.
Compute the correct $E_{cell}^{\circ}$ for this acidic reaction. Anode Fe ( $- 0.44$ ), cathode acidic O₂ ( $+ 1.23$ ): $E_{cell}^{\circ} = 1.23 - (- 0.44) = + 1.67 V, n = 4.$ Why this step? This is the genuine standard driving force for the reaction we actually wrote; it matches the parent note's $+ 1.67$ V.
Write $Q$ from that same balanced reaction. Products on top, reactants on the bottom, each to its coefficient; solids (Fe) and pure water are omitted (activity = 1): $Q = \frac{[ Fe ^{2 +} ] ^{2}}{[ H ^{+} ] ^{4} P _{O_{2}}} = \frac{( 1 0 ^{- 6} ) ^{2}}{( 1 0 ^{- 8} ) ^{4} ( 0.21 )} .$ Why this step? $Q$ measures how far the mixture sits from the 1-molar / 1-atm standard state; only species whose amounts vary appear.
Crunch the powers: numerator $= 1 0^{- 12}$ ; denominator $= 1 0^{- 32} \times 0.21 = 2.1 \times 1 0^{- 33}$ . So $Q = 1 0^{- 12} / (2.1 \times 1 0^{- 33}) \approx 4.76 \times 1 0^{20}$ . Why this step? Subtracting exponents ( $- 12 - (- 33) = 21$ ) is the whole trick with scientific notation — a huge $Q$ means we sit far to the "products-favoured" side.
Apply Nernst: $lo g_{10} (4.76 \times 1 0^{20}) \approx 20.68$ , correction $= \frac{0.059}{4} \times 20.68 = 0.01475 \times 20.68 \approx 0.305$ . $E_{cell} = 1.67 - 0.305 \approx + 1.37 V .$ Why this step? A big positive $lo g_{10} Q$ subtracts from $E^{\circ}$ — building up products or starving reactants (scarce $H^{+}$ ) always weakens the drive, exactly Le Chatelier's logic.
Interpret: even with the Nernst penalty, $\approx + 1.37$ V is strongly positive → the seawater pipe corrodes readily.

The figure below plots $E_{cell}$ against $lo g_{10} Q$ as a straight line sloping down. The red square marks this seawater point; notice it is still well above the dashed $E = 0$ line, so corrosion is alive. The red dot far to the right shows where the line would finally cross zero (corrosion halts) — an enormous $Q$ we never reach here.

Verify: Still $> 0$ → corrosion proceeds in mildly basic seawater, matching real rusting ships. ✓

Worked example Example 4 — Cell D: the degenerate tie (

$E_{cell}^{\circ} \approx 0$ ) Two identical iron plates, same metal, in the same electrolyte — but one is buried in oxygen-poor mud, the other sits in oxygen-rich water. Standard potentials are identical. Does anything corrode?

Forecast: If both electrodes are the same metal, $E_{cell}^{\circ} = 0$ . Naïvely, nothing should happen. Is that right?

Standard picture: $E_{cell}^{\circ} = E_{Fe}^{\circ} - E_{Fe}^{\circ} = 0$ . Why this step? Same electrode both sides → the standard driving force cancels exactly.
But conditions differ! Both sides run the same oxygen cathode $O_{2} + 2 H_{2} O + 4 e^{-} \to 4 OH^{-}$ , so its $Q$ contains $P_{O_{2}}$ (a reactant, on the bottom). With two different $P_{O_{2}}$ values we get a concentration cell. Nernst revives a voltage: $E_{cell} = - \frac{0.059}{n} lo g_{10} \frac{P _{O_{2}, low}}{P _{O_{2}, high}} > 0.$ Why this step? When $E^{\circ} = 0$ the only driving force left is the Nernst concentration term; the electrode sitting in less oxygen has the more negative potential, so nature makes it the anode to even out the imbalance — that's why the term is positive.
Consequence: the oxygen-starved iron (in mud) becomes the anode and corrodes; the oxygen-rich part becomes cathode. This is the famous differential-aeration pitting. Why this step? It explains real, hidden damage — pipes rot underground precisely where oxygen is scarce, not where it is plentiful.

The figure below shows one continuous iron pipe. The dense cloud of black dots on the right marks oxygen-rich water (cathode); the sparse dots on the left mark oxygen-poor mud. The red dot and label pin the pit exactly where oxygen is scarce — the counter-intuitive punchline.

Verify: Take a 100-fold $O_{2}$ ratio, $n = 4$ : $E = - \frac{0.059}{4} lo g_{10} (1/100) = - 0.01475 \times (- 2) = + 0.0295 V$ . Small but positive → the buried region silently rusts. ✓

Worked example Example 5 — Cell E: the wrong-way case

Someone "protects" an iron boat by bolting a copper plate to it in seawater. Compute the cell and explain what actually corrodes.

Forecast: Copper is nobler than iron. When you force two dissimilar metals to touch, which one loses?

Compare $E^{\circ}$ : Cu $+ 0.34$ vs Fe $- 0.44$ . The lower one becomes the anode. Why this step? Lower $E^{\circ}$ = more eager to oxidize = the victim.
So iron is the anode here ( $Fe \to Fe^{2 +} + 2 e^{-}$ ), copper is the cathode. Treat Fe as anode: $E_{cell}^{\circ} = E_{Cu}^{\circ} - E_{Fe}^{\circ} = 0.34 - (- 0.44) = + 0.78 V .$ Why this step? Positive → the galvanic couple is spontaneous, and it's iron being eaten.
Conclusion: copper accelerates iron corrosion. This is the exact opposite of protection — a design blunder called galvanic corrosion. Why this step? It shows that "attach a metal" is not automatically protective; the sign comparison decides everything.

Verify: Contrast with Ex 2: for a protector we needed $E_{anode}^{\circ} < E_{Fe}^{\circ}$ . Copper fails that test ( $+ 0.34 > - 0.44$ ), so predicting iron-as-anode is consistent. ✓

Worked example Example 6 — Cell F: no oxygen — the cathode SWITCHES, it does not vanish

Iron sealed in fully deoxygenated, neutral water ( $P_{O_{2}} \to 0$ ). Does corrosion stop?

Forecast: No dissolved oxygen means the oxygen cathode has nothing to reduce. Guess: does corrosion truly stop, or does something else accept the electrons?

Kill the oxygen route first. For the oxygen cathode, $O_{2}$ is a reactant so $P_{O_{2}}$ sits in the denominator of its $Q$ : $Q_{O_{2}} = \frac{[ OH ^{-} ] ^{4}}{P _{O_{2}} [ H _{2} O ] ^{2}} P_{O_{2}} \to 0 \infty.$ Nernst then gives $E = E^{\circ} - \frac{0.059}{4} lo g_{10} Q_{O_{2}} \to - \infty$ . Why this step? A vanishing reactant blows $Q$ up to infinity, and $lo g_{10} \infty = + \infty$ drags the oxygen-route driving force below zero — so that particular cathode switches off.
But a second cathode is always available: water itself. In the absence of O₂ the reduction that takes over is $2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 OH^{-} (a q) E^{\circ} = - 0.83 V .$ Why this step? Electrons freed by iron must land somewhere; if oxygen is gone, they reduce water to hydrogen gas. This is hydrogen-evolution corrosion, common in oxygen-free boilers and buried anaerobic soils.
Check the new driving force. With Fe as anode ( $- 0.44$ ) and the water cathode ( $- 0.83$ ): $E_{cell}^{\circ} = - 0.83 - (- 0.44) = - 0.39 V .$ Why this step? We reuse the master formula to test the new couple; a negative result means this route is uphill in neutral water, so hydrogen-evolution corrosion of iron is slow/unfavourable at pH 7 — which is why deoxygenated neutral water corrodes iron only sluggishly. It does not stop entirely, and it will run readily as soon as acid lowers the pH (Example 6b).

Verify: $E_{cell}^{\circ} = - 0.83 - (- 0.44) = - 0.39 V$ . Negative but finite (not $- \infty$ ): the electron acceptor has merely switched from O₂ to H₂O, so the conclusion "corrosion shuts off" is wrong — it slows and changes mechanism. ✓

Worked example Example 6b — Cell F′: acidic corrosion, H⁺ is the cathode

The same sealed, oxygen-free iron, but now in acid ( $[H^{+}] = 1 M$ , pH 0). Does iron corrode without any oxygen at all?

Forecast: With plenty of H⁺ and no O₂, guess whether the acid alone can eat the iron.

Pick the cathode that reducing acid supplies: $2 H^{+} + 2 e^{-} \to H_{2}$ , $E^{\circ} = 0.00 V$ . Why this step? In acid, hydrogen ions are the abundant electron acceptors, so H⁺/H₂ (not O₂/OH⁻) is the working cathode.
Combine with the iron anode $Fe \to Fe^{2 +} + 2 e^{-}$ : $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.00 - (- 0.44) = + 0.44 V .$ Why this step? Positive → spontaneous. Overall: $Fe + 2 H^{+} \to Fe^{2 +} + H_{2} ↑$ — the fizzing you see when iron meets dilute acid.
Interpret: acid dissolves iron with no oxygen required; this is why low-pH environments are so corrosive and why acid rain accelerates rusting. Why this step? It closes the loop with Example 6: kill oxygen and add acid, and corrosion comes roaring back through the hydrogen route.

Verify: $E_{cell}^{\circ} = 0.00 - (- 0.44) = + 0.44 V > 0$ , spontaneous — matches the observed vigorous hydrogen bubbling of iron in acid. ✓

Worked example Example 7 — Cell G: Faraday mass-loss word problem

A zinc sacrificial anode delivers a steady protective current of $0.50 A$ for one year (365 days). How many kilograms of zinc are consumed? ( $Zn \to Zn^{2 +} + 2 e^{-}$ , molar mass $65.4 g/mol$ , $F = 96500 C/mol e^{-}$ .)

Forecast: Guess the order of magnitude — grams? Hundreds of grams? Kilograms?

Total charge: $q = I t$ . First get seconds: $t = 365 \times 24 \times 3600 = 3.1536 \times 1 0^{7} s$ . Then $q = 0.50 \times 3.1536 \times 1 0^{7} = 1.5768 \times 1 0^{7} C$ . Why this step? Charge = current × time is the definition of an ampere (coulombs per second).
Moles of electrons: $n_{e} = q / F = 1.5768 \times 1 0^{7} /96500 \approx 163.4 mol e^{-}$ . Why this step? $F$ is the price tag converting coulombs into moles of electrons (Faraday's Laws).
Moles of Zn: each Zn atom gives 2 electrons, so $n_{Zn} = n_{e} /2 \approx 81.7 mol$ . Why this step? Stoichiometry — the "2" in the half-reaction ties electrons to atoms.
Mass: $m = 81.7 \times 65.4 \approx 5343 g \approx 5.34 kg$ . Why this step? Multiplying moles by molar mass converts the count of atoms into a weighable mass.

Verify: Units: $\frac{A \cdot s}{C/mol} = mol e^{-}$ ; divide by 2, times g/mol → grams. ✓ Magnitude of a few kg/year matches real ship-anode replacement schedules. ✓

Worked example Example 8 — Cell H: the exam twist (galvanized scratch)

A galvanized iron sheet (iron coated with zinc, see Passivation) gets scratched down to the bare iron. Does the exposed iron rust at the scratch? Compare with a tin-coated iron sheet scratched the same way.

Forecast: Both coatings are broken. Guess whether the coating still helps — and whether zinc and tin behave the same.

Zinc case: $E_{Zn}^{\circ} = - 0.76 < E_{Fe}^{\circ} = - 0.44$ . Even at a scratch, zinc is the anode; it sacrifices itself and the exposed iron stays cathodic → iron does not rust. Why this step? This is cathodic protection again — the coating protects even where it's gone, because current still flows toward the bare spot.
Tin case: $E_{Sn}^{\circ} = - 0.14 > E_{Fe}^{\circ} = - 0.44$ . Tin is nobler. At a scratch, iron becomes the anode and corrodes — faster than bare iron, because the tin acts as a big cathode. Why this step? Same logic as Example 5 (wrong-way case): the nobler metal drives the base metal's corrosion.
Conclusion: zinc = "active" protection (works even scratched); tin = "barrier" protection (fails badly once breached). This is the classic exam distinction. Why this step? The trap is assuming any coating is equal — the sign of the potential difference decides the fate.

The figure below puts the two cases side by side. In both panels the red box marks the identical scratch. Read the red caption under each: on the left the zinc keeps iron safe; on the right the tin leaves iron to rust faster — same scratch, opposite outcome.

Verify: Zinc–Fe couple $E_{cell}^{\circ}$ with oxygen cathode $= 0.40 - (- 0.76) = + 1.16 V$ (Zn anode ✓). Tin–Fe couple where Fe is anode $= E_{Sn}^{\circ} - E_{Fe}^{\circ} = - 0.14 - (- 0.44) = + 0.30 V > 0$ → iron spontaneously corrodes. ✓

Recall Quick self-test

A protector metal must have $E^{\circ}$ (higher / lower) than the metal it protects ::: lower (more negative) When $O_{2}$ runs out, the cathode does what? ::: it switches to water reduction ( $2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}$ ), not "shuts off" In acid with no oxygen, the cathodic reaction is ::: $2 H^{+} + 2 e^{-} \to H_{2}$ , $E^{\circ} = 0.00$ V Two identical iron electrodes with different $O_{2}$ levels form a ::: concentration (differential-aeration) cell The $lo g$ in the Nernst equation is base ::: ten ( $lo g_{10}$ ) Charge from current and time is computed as ::: $q = I \times t$ Why write $P_{O_{2}}$ not $[O_{2}]$ ? ::: oxygen is a gas; its reactive amount is its partial pressure The oxygen cathode's $E^{\circ}$ is $+ 0.40$ V with OH⁻ but ::: $+ 1.23$ V with H⁺ — match it to your reaction

Related tools used here: Electrochemical Cells, Standard Electrode Potentials, Nernst Equation, Faraday's Laws, Gibs Free Energy, Concentration Cells, Passivation. For a sustainability angle on protection choices, see Green Chemistry.