After introduction of 1.7 ppm of Cl2 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?

Question

After introduction of 1.7 ppm of [latex]Cl_{2}[/latex] 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?

Accepted Answer

Chlorine dissolves in water through the reaction

[latex]Cl_{2}(g)+H_{2}O ightarrow H^{+}+Cl^{-}+HOCl[/latex],

which is the difference of the half reactions

[latex]H_{2}O+Cl^{-} ightarrow H^{+}+HOCl+2e^{-}, \ \ \ \ \ \ \ \ E^{\circ}=+1.49V[/latex],

and

[latex]2Cl^{-} ightarrow Cl_{2}(g)+2e^{-}, \ \ \ \ \ \ \ \ E^{\circ}=+1.36V[/latex].

Applying Eqn (7.14),

[latex]E_{emf}=E^{\circ}+\frac{RT}{nF_{a}}\ln (M^{+n})=E^{\circ}+\frac{0.059}{n}\log (M^{+n}) \ at \ 25^{\circ}C[/latex], (7.14)

[latex]0=(1.49-1.36)+8.314(358)/2(96,500)\ln (H^{+})(Cl^{-})/(Cl_{2})[/latex].

Substituting for the chlorine concentration, [latex](Cl_{2})=1.7 imes 10^{-6}[/latex] yields [latex](H^{+})(Cl^{-})=3.36 imes10^{-10}[/latex] . Thus, there are [latex]1.83 imes10^{-5}[/latex] equivalents per liter for each ion. The equivalent conductance of [latex]H^{+}[/latex] and [latex]Cl^{-}[/latex] is 775 at 85°C [31]. Since [latex] ho=1000/(C\Lambda )[/latex], the sheet resistance, [latex] ho/t_{o}=1000/(C\Lambda t_{o})[/latex]. Substitution yields [latex] ho/t_{o}=1000/1.83 imes 10^{-5}(775)(10^{-5})=7.1 imes 10^{9} \Omega /sq[/latex]

Question 7.1: After introduction of 1.7 ppm of Cl2 into an 85°C/80% RH amb...

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