Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2017 / Paper 1 / Section B / 11 / ii / Solution

On the stated $1{\mathrm{m}}^2$ area, the total electric current is $I=200\mathrm{A}$. The known ohmic polarization is $I{R}_{\Omega }=1.20\mathrm{V}$. Applying the supplied anodic Tafel equation givesAssume the quoted limiting current uses the same cathodic area convention. A steady diffusion-film model gives $c_s/c_b=1-j/j_L=1/41$. The Nernst equation then makes the cathodic concentration overpotential magnitudeRelative to the local surface equilibrium potential, the cathode also needs an activation overpotential. Neglecting anodic concentration polarization, the usual separated-loss accounting isThe printed $b$ is explicitly for copper oxidation, not a separately specified reduction slope. An exchange current density by itself does not determine the whole cathodic current-potential relation. TODO: a unique applied voltage requires a cathodic kinetic law, or an explicit additional relation between anodic and cathodic kinetics.

For a concrete conditional completion, assume one effective two-electron Butler-Volmer equation with ${\alpha }_a+{\alpha }_c=1$, no anodic concentration polarization and reduction proportional to the surface ${\mathrm{Cu}}^{2+}$ concentration. The anodic slope implies ${\alpha }_a=2.303RT/(2Fb)=0.10016$ and ${\alpha }_c=0.89984$. With $f=2F/(RT)$ and cathodic overpotential ${\eta }_c$ referenced to bulk equilibrium,Solving at $j_c=-200\mathrm{A}{\mathrm{m}}^{-2}$ gives ${\eta }_c=-0.12854\mathrm{V}$, including both concentration and activation effects. Therefore this explicitly assumed model gives $V_{\mathrm{app}}=1.20+0.67880+0.12854=2.0073$ V. The $0.04765$ V concentration loss must not be added again. A multistep copper reaction need not obey this effective single-step relation, so 2.0073 V is not uniquely entailed by the printed data.