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CHE/MAE 294 S01 Homework 14 Solution 8.Aug.
restart: readmylib(pchemconstants):
Problem 1
part a
Fe2+ + 2e- -> Fe -0.440 V <== more cathodic (noble) Zn2+ + 2e- -> Zn -0.763 V <== more anodic (reactive)
EMFA := -0.44 - (-0.763); EMFA :=.
part b
Cu2+ + 2e- -> Cu 0.337 V <== more cathodic (noble) Zn2+ + 2e- -> Zn -0.763 V <== more anodic (reactive)
EMFB := 0.337 - (-0.763); EMFB :=1.
part c
Cu+ + e- -> Cu 0.520 V <== more cathodic (noble) Zn2+ + 2e- -> Zn -0.763 V <== more anodic (reactive)
EMFC := 0.52 - (-0.763); EMFC :=1.
Summary
The more positive is the corrosion potential, the greater is the protection against corrosion offered by one of the metals. The corrosion potential increases going from Fe2+ (0.323 V) -> Cu2+ (1.100 V) -> Cu+ (1.283 V). This means, when coupled to a Zn plate, Cu is the least likely metal to corrode, and if it does corode, it is more likely to form Cu2+ than Cu+.
Problem 2
part a
For every 2 e-, 1 atom of Zn corrodes. 1 A of current is 1C/s of charge. The amount of Zn that corrodes will be in g/s. g Zn/s = (1 atom Zn/2 e-)(1 e-/q C)(1 C/s)(mol Zn./No atoms Zn)(g Zn/mol Zn)
mZn := (1/2)(1/q)(1)(1/No)MZn;
mZn :=.5182134300 10 -5^ MZn
MZn := 65.83: mZn; .
0.34 mg Zn/s corrodes when 1 A flows
part b
The answer to this is the same as the answer for part a
Note: If we compare the moles of Fe2+ that react in part a with the moles of Cu+ that react in part b, we would find that 1 mole Fe2+ reacts for every mole of Zn that corrodes, whereas 2 moles of Cu+ react for every 1 mole of Zn that corrodes.
Problem 3
Based on the galvanic series, Zn is more anodic (active) relative to steel. This means, Zn will corrode (oxidize) at the expense of steel. The copper cable serves to provide a path for electron flow to complet the circuit between the oxidizing Zn plate and the non-corroding Fe tank.
