Corrosion

Corrosion is an unintentional galvanic cell. On a real metal surface, microscopic differences in composition, strain, or oxygen access make some patches anodic and others cathodic. The metal itself supplies the electron path between them; a thin film of moisture supplies the ionic path — the two conduction routes any galvanic cell needs, except here anode and cathode sit side by side on one piece of metal rather than in separate half-cells.

where: the anode reaction is the metal loss you actually care about; the cathode reaction merely completes the circuit by consuming the electrons the anode releases. $\text{Fe}^{2+}$ and $\text{OH}^-$ subsequently combine and are oxidised further by dissolved oxygen to hydrated $\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}$, the rust you see — which deposits away from the anodic site, so it never blocks the ongoing metal loss underneath.

Differential aeration decides who is anode. Oxygen reduction is favoured wherever $\text{O}_2$ is more available, so the well-aerated patch of a surface is pulled toward being the cathode and the oxygen-starved patch is forced into the anodic role — the site of actual dissolution. This is why a steel pipe buried in soil often corrodes fastest in its least-aerated stretch, and why a small nick in an otherwise intact coating corrodes violently: a small anode area feeding electrons to a large cathode area concentrates current density at the nick.

Feasibility is read off potentials, exactly as in any cell. The driving force for the couple is the same cell EMF you already compute for any two half-reactions, and its sign tells you whether corrosion (or protection) is thermodynamically favoured:

where: $E^\circ_{cathode}$ and $E^\circ_{anode}$ are standard reduction potentials (V) of the two half-reactions, $n$ is the number of electrons transferred, $F = 96{,}500\ \text{C mol}^{-1}$ is the Faraday constant, and $\Delta G$ (J) is the free-energy change — negative $\Delta G$ (positive $E_{cell}$) means the process is spontaneous. Plotting equilibrium potential against pH for a metal's oxide/hydroxide equilibria (a Pourbaix diagram) extends this same Nernst-equation logic to show which $E$–pH combinations favour the bare metal, its dissolved ions, or a passive oxide film.

Problem. The ship's hull from the sacrificial-anode setup above ($E_{cell} = 0.32\ \text{V}$, $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$) draws a steady protective current of $2.0\ \text{A}$ through its zinc blocks. Find (a) the free-energy change per mole of zinc oxidised, and (b) the mass of zinc consumed over one year of continuous protection ($1\ \text{yr} = 3.156\times10^{7}\ \text{s}$, $M_{\text{Zn}} = 65.4\ \text{g/mol}$, $F = 96{,}500\ \text{C mol}^{-1}$).

Step 1 — Free energy per mole. With $n = 2$, $\Delta G = -nFE_{cell} = -(2)(96{,}500)(0.32) = -61{,}760\ \text{J/mol} \approx -61.8\ \text{kJ/mol}$. The negative sign confirms the zinc-sacrifices-itself reaction is spontaneous, consistent with Step 1 of the derivation.

Step 2 — Total charge passed in one year. $Q = It = (2.0\ \text{A})(3.156\times10^{7}\ \text{s}) = 6.312\times10^{7}\ \text{C}$.

Step 3 — Moles of zinc consumed. Using $n = 2$ for $\text{Zn}^{2+}$: moles $= \dfrac{Q}{nF} = \dfrac{6.312\times10^{7}}{(2)(96{,}500)} = \dfrac{6.312\times10^{7}}{1.93\times10^{5}} \approx 327\ \text{mol}$.

Step 4 — Mass consumed. $m = (327\ \text{mol})(65.4\ \text{g/mol}) \approx 21{,}400\ \text{g}$, so about 21.4 kg of zinc is consumed per year at this protective current — the figure a ship's maintenance schedule is built around.

Competitive stretch

If the current instead varied with season (say $3.0\ \text{A}$ for four winter months and $1.5\ \text{A}$ for eight remaining months), the same formula applies separately to each interval and the masses add, since $Q$ (and hence moles consumed) is simply additive over time — a reminder that $m = ItM/(nF)$ assumes constant $I$ only within each segment you apply it to.

• Assuming more oxygen always means more corrosion at that spot — differential aeration makes the oxygen-poor region the anode, the actual site of metal loss, while the oxygen-rich region merely hosts the cathodic reaction.
• Confusing cathodic protection with a physical coating — cathodic protection works by forcing the metal to be thermodynamically the cathode ($E_{cell}$ and $\Delta G$ shift), not by blocking the electrolyte path the way paint or galvanising does.
• Forgetting the small-anode/large-cathode effect — a tiny exposed anodic area (a scratch in a coating) feeding a large cathodic area concentrates current density there, corroding the scratch far faster than uniform bare metal would.
• Dropping $n$ or using the wrong ionic charge in $m = ItM/(nF)$ — zinc dissolves as $\text{Zn}^{2+}$ ($n=2$), and using $n=1$ would double the predicted mass, understating how much anode metal a given current actually consumes.

Ask the tutor: if the jump from cell EMF to anode consumption mass in the derivation feels like a leap, ask the AI tutor to swap in a magnesium anode and a different current so you can rebuild $m = ItM/(nF)$ from scratch yourself.