Question 4 of 7: Galvanic Concentration Cell and Carbon Diffusion in Steel
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — May 2018. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Notes on the paper state that
any five questions constitute a complete paper and only the first five questions appearing in the
answer book are marked, with all questions of equal value. All seven questions are solved below
for completeness.
Reference texts: Callister & Rethwisch, Materials Science and
Engineering: An Introduction, 9th ed. (crystal structure and packing, polymer molecular
weight, cold work and annealing, corrosion and diffusion, composites, ceramic glasses).
Question 4: Galvanic Concentration Cell and Carbon Diffusion in Steel (20 marks)
Given. (a) $E^o_{Cu^{2+}/Cu}=+0.337$ V; concentrations $0.03$ M and
$0.002$ M; $n=2$ electrons transferred; Nernst equation (page 1)
$E=E_o+\frac{0.0592}{n}\log(C_{ion})$. (b) One carbon atom per $40$ unit cells at the surface, one
per $50$ unit cells at $x=2$ mm; FCC austenite $a_0=0.365$ nm;
$D=3\times10^{-11}$ m$^2$/s at $1000^\circ$C.
Find. (a)(i) Which electrode corrodes. (ii) The open-circuit potential
difference. (b) Carbon atoms crossing each unit cell per minute.
Approach
(a) In a concentration cell of the same metal/ion couple, $E^o$ cancels between the two
half-cells and only the Nernst concentration term survives, so the electrode facing the more
dilute solution has the lower reduction potential and becomes the anode (corrodes). (b) Fick's
first law, $J=-D\,\Delta c/\Delta x$, gives the steady-state flux from the given concentration
gradient; multiplying by the cross-sectional area of one unit-cell face converts the areal flux to
an atoms-per-unit-cell rate.
(a)(i) Which end corrodes. The half-cell reduction potential is
$E=E^o+\frac{0.0592}{2}\log[Cu^{2+}]$. Since $\log(0.002)<\log(0.03)$, the dilute
($0.002$ M) electrode has the lower reduction potential, so it is the less noble
(anodic) end and it is the one that corrodes — a concentration cell always
drives the metal in contact with the more dilute solution to dissolve, in the direction that would
tend to equalize the two concentrations.
(a)(ii) Potential difference. The cell EMF is the reduction-potential
difference between the two half-cells; $E^o$ cancels because both ends are the same Cu/Cu$^{2+}$
couple:
$$E_{cell}=\frac{0.0592}{2}\log\!\left(\frac{0.03}{0.002}\right)=\frac{0.0592}{2}\log(15)$$
$$\boxed{E_{cell}\approx0.0348\ \text{V}\approx34.8\ \text{mV}}$$
(b) Carbon concentrations. With $n_{cell}$ carbon atoms per unit cell and
cell volume $a_0^3$, concentration is $c=n_{cell}/a_0^3$. With $a_0=3.65\times10^{-10}$ m,
$a_0^3=4.865\times10^{-29}$ m$^3$:
$$c_{surf}=\frac{1/40}{a_0^3}\approx5.14\times10^{26}\ \text{atoms/m}^3,\qquad
c_{2mm}=\frac{1/50}{a_0^3}\approx4.11\times10^{26}\ \text{atoms/m}^3$$
(b) Flux and atoms per unit cell per minute. Fick's first law over
$\Delta x=2\times10^{-3}$ m:
$$J=D\,\frac{c_{surf}-c_{2mm}}{\Delta x}=(3\times10^{-11})\frac{(5.14-4.11)\times10^{26}}{2\times10^{-3}}\approx1.54\times10^{18}\ \text{atoms/(m}^2\text{s)}$$
Multiplying by one unit-cell face area $a_0^2=1.332\times10^{-19}$ m$^2$ and converting to
minutes:
$$\text{rate}=J\,a_0^2\times60=(1.54\times10^{18})(1.332\times10^{-19})(60)$$
$$\boxed{\text{rate}\approx12.3\ \text{carbon atoms per unit cell per minute}}$$
