Question 7.1: After introduction of 1.7 ppm of Cl2 into an 85°C/80% RH amb...
After introduction of 1.7 ppm of Cl_{2} into an 85°C/80% RH ambient, the corrosion leakage current across metal stripes on a ceramic substrate increased. If the moisture layer is 100 nm thick, what is the electrolyte sheet resistance?
Chlorine dissolves in water through the reaction
Cl_{2}(g)+H_{2}O\rightarrow H^{+}+Cl^{-}+HOCl,
which is the difference of the half reactions
H_{2}O+Cl^{-}\rightarrow H^{+}+HOCl+2e^{-}, \ \ \ \ \ \ \ \ E^{\circ}=+1.49V,
and
2Cl^{-}\rightarrow Cl_{2}(g)+2e^{-}, \ \ \ \ \ \ \ \ E^{\circ}=+1.36V.
Applying Eqn (7.14),
E_{emf}=E^{\circ}+\frac{RT}{nF_{a}}\ln (M^{+n})=E^{\circ}+\frac{0.059}{n}\log (M^{+n}) \ at \ 25^{\circ}C, (7.14)
0=(1.49-1.36)+8.314(358)/2(96,500)\ln (H^{+})(Cl^{-})/(Cl_{2}).
Substituting for the chlorine concentration, (Cl_{2})=1.7\times 10^{-6} yields (H^{+})(Cl^{-})=3.36\times10^{-10} . Thus, there are 1.83\times10^{-5} equivalents per liter for each ion. The equivalent conductance of H^{+} and Cl^{-} is 775 at 85°C [31]. Since \rho=1000/(C\Lambda ), the sheet resistance, \rho/t_{o}=1000/(C\Lambda t_{o}). Substitution yields \rho/t_{o}=1000/1.83 \times 10^{-5}(775)(10^{-5})=7.1 \times 10^{9} \Omega /sq