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...

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.

The reaction you suggested, O2(g) + 4H+ + 4e- -> 2H2O, is the reduction half-reaction that occurs during the corrosion of a galvanized nail exposed to rain. This half-reaction involves the reduction of oxygen gas (O2) to water (H2O) with the gain of 4 electrons.

Zn(s) -> Zn2+(aq) + 2e- is the oxidation half-reaction that occurs during the corrosion of the galvanized nail. In this reaction, solid zinc (Zn) is oxidized, losing 2 electrons to form zinc ions in solution.

The overall reaction is the combination of these two half-reactions:

Zn(s) + O2(g) + 4H+ + 4e- -> Zn2+(aq) + 2e- + 2H2O

Simplifying the equation, we have:

Zn(s) + O2(g) + 4H+ -> Zn2+(aq) + 2H2O

Regarding the potential, it is important to note that the potential values are specific to the half-reactions and can vary depending on the conditions. The potential values indicate the tendency of a half-reaction to occur spontaneously. In the case of corrosion, a positive potential is favorable for the overall reaction.

In your suggested half-reaction, the potential of +1.23V for the reduction of oxygen gas is correct. However, the potential of +0.40V you mentioned for the alternative half-reaction involving the formation of hydroxide ions is also correct. Both reactions can occur, but the reduction of oxygen gas to form water is the more favorable and predominant reaction during the corrosion of a galvanized nail.

As for the acidity, rainwater is slightly acidic due to the presence of dissolved carbon dioxide (CO2), which forms carbonic acid (H2CO3) when it reacts with water. However, in this case, the reduction of oxygen gas to water is still the primary reaction that takes place during the corrosion of the galvanized nail in rain.

You're on the right track with considering the reaction between zinc and oxygen in the presence of rain. However, there are a few things to clarify.

When a galvanized nail (composed of zinc) is exposed to rain, the oxidation and reduction half-reactions involved in the corrosion process can be as follows:

Oxidation half-reaction: Zn(s) -> Zn2+(aq) + 2e-

Reduction half-reaction: 2H2O(l) + 2e- -> H2(g) + 2OH-(aq)

Let's break down why these half-reactions are chosen over the ones you have proposed.

Regarding the oxidation half-reaction, you correctly identified that the zinc metal (Zn) is oxidized to form zinc ions (Zn2+) and release two electrons (2e-). This reaction occurs at the anode of the corrosion cell.

For the reduction half-reaction, it involves the reduction of water molecules (H2O) and the formation of hydrogen gas (H2) and hydroxide ions (OH-). This reaction occurs at the cathode of the corrosion cell.

You mentioned that rain is slightly acidic, and normally, one might expect the reduction of oxygen (O2) to form water (H2O) in the presence of protons (H+). However, in the case of galvanized zinc, the iron or steel substrate underneath the zinc also plays a role.

In the presence of both zinc and iron/steel, the corrosion process on the galvanized nail is typically controlled by the reduction half-reaction involving water reduction and the formation of hydroxide ions (OH-). This is because the corrosion reaction on the zinc surface is generally faster than on the iron/steel surface.

The reduction half-reaction involving the formation of hydroxide ions (OH-) instead of water (H2O) is chosen because it more accurately represents the corrosion process observed in galvanized nails.

It's important to note that the potentials you provided (0.76V and 1.23V) are standard reduction potentials, which indicate the tendency of a species to be reduced. The overall potential of the corrosion reaction depends on the specific conditions and concentrations of the reactants present. In this case, the potential of the corrosion reaction is determined by the potential difference between the oxidation and reduction half-reactions mentioned above.

Therefore, the correct half-reactions for a corroding galvanized nail exposed to rain are:

Oxidation half-reaction: Zn(s) -> Zn2+(aq) + 2e- (potential: +0.76V)

Reduction half-reaction: 2H2O(l) + 2e- -> H2(g) + 2OH-(aq) (potential: +0.40V)