LoganJohnston

Yes it is I have since been explained the answer thusly: "Corrosion is the result of a redox reaction. Shown below is the reduction half-reaction. Now if we add more hydroxide ions, the reaction will shift towards the left, correct? This will impede the reduction half-reaction – it will not occur to any great extent. Hence there will be very little corrosion." We were on the right track, although something seems wrong with the way this question was worded. Or perhaps something was wrong with my interpretation. Oh well, something to think about I suppose. Thank you all for your input.

I agree, I worked that out on NYE. So to summarize: the reaction of iron in the presence of water and O2 is spontaneous, yes. In the presence of hydroxide, there is no O2 to react with so no reaction occurs, which explains the lack of corrosion seen in the earlier lab the question references. I understand all of that just fine now, but I will have to email my teacher as I don't understand why the question states the reaction is spontaneous. In answer to my original question of explaining the non-reaction with the provided half-reaction, could it be that the concentration of OH-(aq) is much higher than the concentration of O2(g), thus no O2(g) is present to react with the iron nail?

using standard E values O2(g) + 2 H2O(l) + 4 e– ⇌ 4 OH– (aq) ... E°+0.40 Fe(s) ⇌ Fe2+(aq) + 2 e– ... E°+ 0.45 And this sentence from the question : "This observation is difficult to explain from an electrochemistry perspective since electrochemistry principles predict a spontaneous reaction that should cause corrosion."

The question asks: Recall that you observed very little corrosion occurring on the iron nail immersed in NaOH(aq) solution. This observation is difficult to explain from an electrochemistry perspective since electrochemistry principles predict a spontaneous reaction that should cause corrosion. Explain why there was no corrosion on this nail from an equilibrium perspective using the half-reaction shown below: O2(g) + 2 H2O(l) + 4 e- --> 4 OH-(aq) I understand that the reaction is spontaneous between NaOH(aq) and Fe(s), yet no visible precipitate forms. From an equilibrium perspective though, I am truly stumped. It obviously has something to do with Le Châtelier's Principle, but since the solution is a condensed state I'm not sure how to apply the principle. Any thoughts?