Write the oxidation and reduction half-reactions of a corroding galvanized nail exposed to rain.
The question says "galvanized" so its zinc.
Zn(s) -> Zn2+(aq) + 2e- potential: +0.76V
rain is slightly acidic, so i thought it would be
O2(g) + 4H+ + 4e- -> 2H2O potential: +1.23V
the total potential is +1.99V so it should be spontaneous
but the answer in the book tells me that it should be
O2(g) + H2O(l) + 2e- -> 2OH-(aq) with potential +0.40V
I don't understand, first, shouldn't the reactant be acidic? and this one has less potential...
2 answers
I can't answer your question but I may be able to offer some insight. The equation you wrote is, as you say, an acid solution. Remember that the standard potential is for 1 M H^+ and acid rain isn't close to that. The equation offered up as the answer by the book is for an alkaline solution; again, it's for a 1 M solution of base and rain isn't close to that either. I looked in my copy of General Chemistry by Whitten, Davis, and Peck and they talk about corrosion of an iron nail. They write the same expression as offered in the book [perhaps we are looking at the same book :-)] although they use Fe and not Zn since their nail is not galvanized. There is another equation for O2/H2O in neutral solution in which the potential is 0.828. One possible explanation could be that, when corrected for concentration, that the E value is closer to that of the equation offered than to the 1.23. If this is an advanced course, that won't cut it but if it is for freshman chemistry, it might get you by. Personally, I think the rain water is slightly acidic (environmentalists aside---just from CO2 if nothing else) and that's the equation that should be used. However, I also think it should be corrected for H^+ concentration.
I forgot to mention that the equation you wrote for the answer provided by the book is not balanced. 3 O atoms on left and 2 on right and the charge doesn't balance either.