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 separated-loss accounting isTo calculate that activation term, adopt a single effective two-electron Butler-Volmer equation with constant transfer coefficients satisfying ${\alpha }_a+{\alpha }_c=1$, no anodic concentration polarization and reduction proportional to the surface ${\mathrm{Cu}}^{2+}$ concentration. The supplied anodic Tafel slope then determines the reverse slope throughWith $f=2F/(RT)$ and cathodic overpotential ${\eta }_c$ referenced to bulk equilibrium,Solving at $j_c=-200\mathrm{A}{\mathrm{m}}^{-2}$ gives ${\eta }_c\simeq -0.12854\mathrm{V}$, including both concentration and activation effects. ThusThe $0.04765$ V concentration loss must not be added again. The kinetic assumption matters: the supplied slope is for copper oxidation, and an exchange current density alone does not specify every cathodic current-potential relation. A multistep copper reaction need not obey the effective single-step relation used for this estimate.