Chemistry Labs

Problem 2

Bridge between Denmark and Sweden. The combined tunnel and bridge between Copenhagen and Malmö uses concrete and steel; this problem concerns reactions in their production and degradation. Concrete is made from cement, water, sand and stones. Cement consists mainly of calcium silicates and aluminates; in the final step some gypsum, CaSOX4 ⋅ 2 HX2O\ce{CaSO4.2H2O}, is added to improve hardening. Elevated temperatures may form unwanted hemihydrate: CaSOX4 ⋅ 2 HX2O(s)→CaSOX4 ⋅ 12 HX2O(s)+32 HX2O(g)\ce{CaSO4.2H2O(s) -> CaSO4.1/2H2O(s) + 3/2 H2O(g)}. Thermodynamic data at 25 ∘C25\ ^\circ\mathrm{C}, 1.00 bar — CaSOX4 ⋅ 2 HX2O(s)\ce{CaSO4.2H2O(s)}: ΔHf∘=−2021.0\Delta H_f^\circ = -2021.0 kJ mol−1^{-1}, S∘=194.0S^\circ = 194.0 J K−1^{-1} mol−1^{-1}; CaSOX4 ⋅ 12 HX2O(s)\ce{CaSO4.1/2H2O(s)}: −1575.0-1575.0 kJ mol−1^{-1}, 130.5130.5 J K−1^{-1} mol−1^{-1}; HX2O(g)\ce{H2O(g)}: −241.8-241.8 kJ mol−1^{-1}, 188.6188.6 J K−1^{-1} mol−1^{-1}. R=8.314R = 8.314 J mol−1^{-1} K−1^{-1}; 0 ∘C=273.150\ ^\circ\mathrm{C} = 273.15 K. (2.1) Calculate ΔH∘\Delta H^\circ for the transformation of 1.00 kg of CaSOX4 ⋅ 2 HX2O(s)\ce{CaSO4.2H2O(s)} into hemihydrate; is it endothermic or exothermic? (2.2) Calculate the equilibrium water-vapour pressure (bar) in a closed vessel containing CaSOX4 ⋅ 2 HX2O(s)\ce{CaSO4.2H2O(s)}, CaSOX4 ⋅ 12 HX2O(s)\ce{CaSO4.1/2H2O(s)} and HX2O(g)\ce{H2O(g)} at 25 ∘C25\ ^\circ\mathrm{C}. (2.3) Calculate the temperature at which the equilibrium water-vapour pressure is 1.00 bar, assuming ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ are temperature-independent. Corrosion of iron involves the electrode reactions Fe(s)→FeX2+(aq)+2 eX−\ce{Fe(s) -> Fe^{2+}(aq) + 2e-} and OX2(g)+2 HX2O(l)+4 eX−→4 OHX−(aq)\ce{O2(g) + 2H2O(l) + 4e- -> 4OH-(aq)}. Consider the cell Fe(s)∣FeX2+(aq)∣∣OHX−(aq), OX2(g)∣Pt(s)\ce{Fe(s)|Fe^{2+}(aq)||OH-(aq),O2(g)|Pt(s)} at 25 ∘C25\ ^\circ\mathrm{C} with E∘(FeX2+/Fe)=−0.44E^\circ(\ce{Fe^{2+}/Fe}) = -0.44 V and E∘(OX2/OHX−)=+0.40E^\circ(\ce{O2/OH-}) = +0.40 V; RTln⁡10/F=0.05916RT\ln10/F = 0.05916 V, F=96485F = 96485 C mol−1^{-1}. (2.4) Calculate the standard EMF at 25 ∘C25\ ^\circ\mathrm{C}. (2.5) Write the overall discharge reaction. (2.6) Calculate the equilibrium constant at 25 ∘C25\ ^\circ\mathrm{C}. (2.7) The cell runs 24 h at a constant 0.12 A under standard conditions; calculate the mass of Fe converted to FeX2+\ce{Fe^{2+}}. (2.8) Calculate EE for [FeX2+]=0.015[\ce{Fe^{2+}}] = 0.015 M, pH of the right half-cell 9.00, p(OX2)=0.700p(\ce{O2}) = 0.700 bar.
Step 5 of 5: Corroded mass and cell EMF
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Q=It=0.12×86400=10368 C;m(Fe)=Q2F×55.85=3.0 g;E=0.84−0.059164log⁡0.015 (10−5)40.700=1.19 VQ = It = 0.12\times86400 = 10368\ \text{C};\quad m(\ce{Fe}) = \frac{Q}{2F}\times55.85 = 3.0\ \text{g};\quad E = 0.84 - \frac{0.05916}{4}\log\frac{0.015\,(10^{-5})^4}{0.700} = 1.19\ \text{V}
Analysis

The charge Q=10368Q = 10368 C carries Q/F=0.1075Q/F = 0.1075 mol of electrons; since each Fe loses 2 e−^-, m(Fe)=0.1075/2×55.85=3.0m(\ce{Fe}) = 0.1075/2\times55.85 = 3.0 g. For (2.8), pH 9.00 gives [OHX−]=10−5[\ce{OH-}] = 10^{-5} M, and the Nernst equation E=E∘−0.059164log⁡[FeX2+][OHX−]4p(OX2)E = E^\circ - \frac{0.05916}{4}\log\frac{[\ce{Fe^{2+}}][\ce{OH-}]^4}{p(\ce{O2})} yields 1.19 V.

Common pitfall. The reaction quotient uses both [FeX2+][\ce{Fe^{2+}}] and [OHX−]4[\ce{OH-}]^4; do not forget that pH fixes [OHX−][\ce{OH-}], not [HX+][\ce{H+}], on the right-hand side.