Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 1 / Section A(iii) / 6

The more negative standard reduction potential identifies iron as the species more readily oxidized. Combining $\mathrm{Fe}\to \mathrm{F}{\mathrm{e}}^{2+}+2{\mathrm{e}}^{-}$ with $\mathrm{S}{\mathrm{n}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Sn}$ gives the spontaneous standard galvanic cell reactionThe electron balance is two per reaction, and the activities of the pure metals are unity.

For a galvanic cell, subtract the standard anode reduction potential from the standard cathode reduction potential:A reduction potential is not multiplied by the number of transferred electrons.

Iron is the negative anode and tin is the positive cathode of this spontaneous galvanic cell. electrons travel through the external circuit from iron to tin. An intact tin coating protects iron mainly by physically excluding the corrosive medium. If a defect exposes iron while the metals remain electrically connected, tin is not a sacrificial coating: the exposed iron can become the anodic corrosion site. A small exposed iron area coupled to a large cathodic tin area can be especially damaging; this distinction follows from the electrode-potential order rather than the appearance of the coating.

With $n=2$, the standard reaction Gibbs free energy and dimensionless equilibrium constant areFor the reaction as written, $Q=a_{\mathrm{F}{\mathrm{e}}^{2+}}/a_{\mathrm{S}{\mathrm{n}}^{2+}}$, and the Nernst equation gives $E=E^{\circ }-(RT/2F)\ln Q$. Consequently zero cell potential requires $[\mathrm{F}{\mathrm{e}}^{2+}]/[\mathrm{S}{\mathrm{n}}^{2+}]=K\simeq 1.93\times 10^{10}$ under the specified unit-activity-coefficient assumption. Reversing that ratio reverses the conclusion. This mathematical ratio does not by itself guarantee experimentally attainable concentrations without precipitation or other chemistry.