Question 5 of 8: Cu Concentration Cell and Carbon Diffusion in FCC Iron

Question 5: Cu Concentration Cell and Carbon Diffusion in FCC Iron (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) $E^\circ_{Cu}=+0.337$ V, $n=2$; $C_1=0.03\ M$, $C_2=0.002\ M$ Cu$^{2+}$. (b) Surface concentration 1 C atom / 40 unit cells; concentration at $x=2$ mm, 1 C atom / 50 unit cells; $D=3\times10^{-11}\ \text{m}^2/\text{s}$; FCC iron, $a_0=0.365$ nm.

Find. (a)(i) Which end corrodes. (a)(ii) Potential difference. (b) Carbon atoms diffusing through each unit cell per minute.

Approach

Part (a) is a concentration cell: with identical electrode material at both ends, the Nernst equation gives a potential difference driven purely by the ion-concentration difference, and thermodynamics dictates that the electrode in the more dilute solution is the one that must oxidize (corrode) to restore equilibrium. Part (b) applies Fick's first law directly, converting the given "atoms per unit cell" concentrations into a true volumetric concentration using the FCC unit-cell volume, then converts the resulting areal flux into a per-unit-cell diffusion rate.

1. (a)(i) Which end corrodes. In a concentration cell with identical metal at both electrodes, the electrode facing the more dilute ion solution has a lower Nernst (reduction) potential, making it the anode: it must dissolve (oxidize, $\text{Cu}\to \text{Cu}^{2+}+2e^-$) to raise its local ion concentration back toward equilibrium. Here the $0.002\ M$ end is dilute, so the end in the 0.002 M solution corrodes; the 0.03 M end is the cathode (protected).
2. (a)(ii) Potential difference. By the Nernst equation applied to each half-cell and subtracting (the standard potentials cancel since both ends are copper), $$\Delta E=\frac{0.0592}{n}\log\!\left(\frac{C_{hi}}{C_{lo}}\right) =\frac{0.0592}{2}\log\!\left(\frac{0.03}{0.002}\right)=\frac{0.0592}{2}\log(15)=\boxed{34.8\ \text{mV}}.$$
3. (b) Convert atoms-per-unit-cell to a volumetric concentration. FCC unit-cell volume $V_{cell}=a_0^3=(0.365\times10^{-9})^3=4.865\times10^{-29}\ \text{m}^3$. Surface concentration $$c_1=\frac{1\ \text{atom}}{40\,V_{cell}}=5.14\times10^{26}\ \text{atoms/m}^3,$$ and at 2 mm depth $$c_2=\frac{1\ \text{atom}}{50\,V_{cell}}=4.11\times10^{26}\ \text{atoms/m}^3.$$
4. Fick's first law flux. With $\Delta x=2\ \text{mm}=2\times10^{-3}$ m, $$J=D\,\frac{c_1-c_2}{\Delta x}=(3\times10^{-11})\times\frac{(5.14-4.11)\times10^{26}}{2\times10^{-3}} =1.54\times10^{18}\ \text{atoms/(m}^2\cdot\text{s)}.$$
5. Convert to atoms per unit cell per minute. The flux is atoms crossing a unit area per second; scaling by one unit-cell cross-sectional face area $a_0^2$ and then by 60 s/min, $$\dot n_{cell}=J\,a_0^2\times60=(1.54\times10^{18})\times(0.365\times10^{-9})^2\times60 =\boxed{12.3\ \text{C atoms per unit cell per minute}}.$$