Mixed-Potential Theory — Pb-Sn Solder: 21-Mat-A7 Environmental Degradation of Materials, May 2018, Q5

Question 5: Mixed-Potential Theory — Pb-Sn Solder (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. Four half-reactions with Tafel kinetics referenced to their own equilibrium potentials; two of the cathodic branches (Pb, Sn reduction back to metal, and O$_2$ reduction) carry a stated diffusion-limiting current $i_{LIM}$, while H$^+$ reduction has none in the given range.

Given data — electrochemical kinetic parameters
Reaction	$E_0$ (V vs SHE)	$i_0$ (A/cm²)	$\beta$ (mV/decade)	$i_{LIM}$ (A/cm²)
$Pb^{2+}\leftrightarrow Pb$	−0.8	$1\times10^{-8}$	80	0.7
$Sn^{2+}\leftrightarrow Sn$	−0.9	$8\times10^{-6}$	100	0.1
$H^+\rightarrow H_2$	0.0	$3\times10^{-8}$	200	— (none)
$O_2\rightarrow H_2O$	1.229	$1\times10^{-7}$	240	0.002

Find. The overall corrosion potential $E_{corr}$ and corrosion current density $i_{corr}$ of the Pb-Sn couple, and the individual partial current of each of the four reactions at that potential.

Approach. Pb and Sn (the solder metals, at very active $E_0$) act as the couple's total anode via their own anodic Tafel branches $i_a=i_0\,10^{(E-E_0)/\beta}$ (no diffusion limit on solid-metal dissolution). H$^+$ and O$_2$ (at far more noble $E_0$) act as the couple's total cathode via their reduction branches, each under mixed activation/diffusion control where a limit is given: $1/i_c = 1/i_{0}10^{-(E-E_0)/\beta} + 1/i_{LIM}$. Mixed-potential theory sets $E_{corr}$ where the summed anodic curve crosses the summed cathodic curve; the intersection is read from the plotted total lines exactly as the graph paper instructs.

Computed mixed-potential polarization diagram for the Pb-Sn solder couple: individual anodic (solid) and cathodic (dashed) branches for Pb, Sn, H+, and O2, plus the summed total-anodic and total-cathodic lines (heavy) whose intersection gives Ecorr and icorr.

(a)/(b) Plot each reaction and sum to the total anode/cathode lines. The figure above plots all four individual branches (thin lines: solid = anodic branch, dashed = cathodic branch, one colour per reaction) together with the two heavy summed curves — total anodic (Pb dissolution $+$ Sn dissolution) and total cathodic (H$^+$ reduction $+$ O$_2$ reduction, the O$_2$ branch bending over into its diffusion plateau near $i_{LIM,O_2}=2\times10^{-3}$ A/cm$^2$). These two heavy curves are the "overall anode and cathode lines" mixed-potential theory asks for.
(c) Read $E_{corr}$/$i_{corr}$ from the intersection of the two heavy lines. $$\boxed{E_{corr}\approx -0.66\ \text{V vs SHE}, \quad i_{corr}\approx 2.06\times10^{-3}\ \text{A/cm}^2\ (2060\ \mu\text{A/cm}^2)}$$ The intersection is pulled right down into the active-metal region — only about 140 mV above Pb's $E_0=-0.8$ V, and nowhere near the two cathodic reactions' own equilibrium potentials — and it is tin that puts it there. Sn's exchange current density is nearly three orders of magnitude larger than Pb's, so Sn's anodic branch dominates the total anodic curve and fixes the crossing at $-0.66$ V; lead alone, against the same two cathodic reactions, would sit at about $-0.38$ V. Tin therefore carries essentially all of the corrosion current even though lead has the more noble $E_0$ of the two.
(d) Partial current of each reaction at $E_{corr}=-0.66$ V. Evaluate each reaction's own anodic and cathodic branch at $E_{corr}$ and keep whichever is dominant (the other branch of that same couple is negligible at this potential, confirming the anode/cathode assumption in the Approach): $$i_{a,Pb}=i_{0,Pb}\,10^{(E_{corr}-E_{0,Pb})/\beta_{Pb}} = (10^{-8})\,10^{(0.14)/0.08} \approx 5.8\times10^{-7}\ \text{A/cm}^2$$ $$i_{a,Sn}=i_{0,Sn}\,10^{(E_{corr}-E_{0,Sn})/\beta_{Sn}} = (8\times10^{-6})\,10^{(0.24)/0.10} \approx 2.06\times10^{-3}\ \text{A/cm}^2$$ $$i_{c,H}=i_{0,H}\,10^{-(E_{corr}-E_{0,H})/\beta_H} = (3\times10^{-8})\,10^{(0.66)/0.20} \approx 5.9\times10^{-5}\ \text{A/cm}^2$$ $$i_{c,O_2}\approx i_{LIM,O_2} = 2.0\times10^{-3}\ \text{A/cm}^2\ \text{(fully diffusion-limited at this potential)}$$

Final results — Question 5
Reaction	$i$ at $E_{corr}$ (A/cm²)	Anode or cathode?
$Pb^{2+}\leftrightarrow Pb$	$5.8\times10^{-7}$	Anode (negligible — Pb barely corrodes)
$Sn^{2+}\leftrightarrow Sn$	$2.06\times10^{-3}$	Anode (dominant — essentially all the corrosion current)
$H^{+}\rightarrow H_2$	$5.9\times10^{-5}$	Cathode (minor)
$O_2\rightarrow H_2O$	$2.0\times10^{-3}$	Cathode (dominant, diffusion-limited)
Overall	$E_{corr}=-0.66$ V, $i_{corr}=2.06\times10^{-3}$ A/cm$^2$	—

Check

Read graphically from the graph paper supplied with the exam, $E_{corr}$/$i_{corr}$ would ordinarily carry $\pm1$ small-division reading uncertainty (of order $\pm0.02$ V, $\pm10\%$ on $i$); the values above are the exact numerical intersection of the given Tafel/limiting-current expressions, i.e. the value the graphical construction is targeting.

Concept & Theory

Topic. Mixed-potential theory applied to a galvanic couple of more than two electrode reactions, with diffusion-limited cathodic partners.

Key relations. Anodic Tafel $i_a=i_0\,10^{(E-E_0)/\beta_a}$; cathodic Tafel with mass-transport limit $1/i_c=1/(i_0\,10^{-(E-E_0)/\beta_c})+1/i_{LIM}$; mixed-potential condition $\sum i_{a,k}(E_{corr})=\sum i_{c,j}(E_{corr})$ (Wagner-Traud); charge conservation at the corroding surface (no net current flows to the external environment).

Why this works. A corroding surface carrying several electrode reactions simultaneously must be at a single, common potential (Ohm's law inside a highly conductive metal makes any potential gradient across the surface negligible), so every reaction is forced away from its own equilibrium by whatever amount is needed to make the total anodic and total cathodic currents balance. Sn dominates the anodic side here not because it is more "active" in a simple sense but because its exchange current density is so much larger that its curve out-produces Pb's by orders of magnitude at any shared potential — consistent with the well-known metallurgical fact that Sn, not Pb, is the component that actually corrodes preferentially in Pb-Sn solder joints.

Common pitfall. Averaging the four $E_0$ values to estimate $E_{corr}$ (ignoring the exchange-current-density and Tafel-slope differences that actually control which reaction dominates), or forgetting that a diffusion-limited cathodic branch plateaus rather than continuing to rise on the Tafel line once $i_{LIM}$ is reached.

Source. Jones, Principles and Prevention of Corrosion, 2nd ed., Ch. 4 (mixed-potential theory, multi-electrode systems); Fontana, Corrosion Engineering, 3rd ed., Ch. 6.

Question 5 of 8: Mixed-Potential Theory — Pb-Sn Solder

Question 5: Mixed-Potential Theory — Pb-Sn Solder (20 marks)