[LoganJohstom]

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)

My thought is to consider the electrons in the half reaction as an indicator of equilibrium. Since the electrons are present on the reactant side, equilibrium favours the reactants and no reaction occurs. When compared to the half reaction for the reduction of iron metal, there are more electrons on the reactant sides than the product , confirming the fact that no reaction occurs.

My work:

O2(g) + 2 H2O(l) + 4 e– ⇌ 4 OH– (aq)
Fe(s) —> 2e- + Fe2+

Fe(s) +O2(g) + 2H2O(l) + 4e- = 2e- + Fe2+ 4OH-(aq)

The equilibrium of charge in this mixture is more towards the reactant side, therefore no reaction will take place. The iron nail will not reduce to form iron oxide (rust).


I'm doing all this work without anyone to talk it out with, so... any thoughts? Am I on the right track or am I forgetting something here?

[mikasaur]

I'm a student too so take my thoughts with a grain of salt.

                           O2(g)  +  2 H2O(l)  +  4 e-  -->  4 OH-(aq)

My thought is to consider the electrons in the half reaction as an indicator of equilibrium. Since the electrons are present on the reactant side, equilibrium favours the reactants and no reaction occurs. When compared to the half reaction for the reduction of iron metal, there are more electrons on the reactant sides than the product , confirming the fact that no reaction occurs.

I'm not sure that this makes sense. Electrons on one side of a reaction or the other do not indicate whether the reactants or products are favorable or not. Take for instance, the reduction of Ag²⁺ compared to the reduction of Li⁺.

[itex]\ce{Ag^2+ + e- -> Ag+}\;\;\;\;\;\;\textrm{E}^{\circ}=+1.98\textrm{ V}[/itex]
[itex]\ce{Li+ + e- -> Li}\;\;\;\;\;\;\textrm{E}^{\circ}=-3.05\textrm{ V}[/itex]

The electrons are on the same side of both reactions, but the high positive E° of Ag²⁺ means that it is strongly oxidizing and so this reaction "tends" toward the forward direction under standard conditions (the Ag²⁺ "likes" being reduced). The opposite could be said of Li⁺.

You don't have them listed here, but I bet you have access to the standard potentials (E°) of your reactions involving Fe and OH^-. The key word being standard. But what happens when you put NaOH in water? How might that affect one or both of the potentials?

[LoganJohstom]

On second look I may have been too liberal with my flipping the Fe half reaction to satisfy my hypothesis.

Speaking of standard potentials, what if the addition of the standard potential for OH- to the standard potential of Fe (reduction) is indicative of reaction?

(OH-(aq))E° = +.40V
(Fe(s))E°     = -.45V

 .40V+(-.45V) = -.05V

Negative E° means the reaction is non-spontaneous. I'm struggling to prove this by equilibrium however...

I'm not sure how to answer your second question: "How might [adding NaOH to water] affect one or both of the potentials?"

[mikasaur]

Hmm, well so we have two half reactions going on:

[itex]\ce{O2(g) + 2H2O + 4e- -> 4OH- (aq)}[/itex]	[itex]\textrm{E}°=0.40 \textrm{ V}[/itex]
[itex]\ce{Fe2+(aq) + 2e- -> Fe(s)}[/itex]	[itex]\textrm{E}°=-0.44 \textrm{ V}[/itex]

You have the right idea that these two are going to interact in some way, but these are both reduction reactions. If they're going to interact from an electrochemical perspective, one species must be donating electrons (oxidation) and the other must be accepting electrons (reduction). So one must be flipped, right? Which one do you think is going to want to flip?

We may be going down the wrong path though, because the question states that from an electrochemical standpoint the corrosion of Fe will be spontaneous (can you see why?).

So like you mentioned, let's focus on equilibrium. Remember Le Châtelier's principle? What might happen to one of our reactions of interest when we add a bunch of NaOH in solution? How might that affect how our two reactions "interact" with each other?

[LoganJohstom]

Fe_(s) would have to show as an oxidation so it would flip, yes.

If Fe flips then the sumE° is positive, so a spontaneous reaction occurs. However, very little corrosion does occur in the experiment. So I can see that, from an electrochemical standpoint, the reaction should appear spontaneous. That's clear now.

Increasing concentration of reactants shifts equilibrium forward to the product side, but that won't work for liquids unless we add higher concentration NaOH(aq).

[LoganJohstom]

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 by considering equilibrium in a half-reaction, 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 for your input mikasaur.

[mikasaur]

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."

This is where I was trying to lead you when I asked what would happen if we added a bunch of NaOH. NaOH is a strong base that completely ionizes so the concentration of OH^- will be very high and will impede the reduction half-reaction, as stated.

What tripped me up a bit is that this is still explainable with electrochemistry. I'm not sure if you've learned about it yet but the Nernst equation helps you to find E from E°. It is the mathematical interpretation of impeding the reduction half-reaction; by greatly increasing our concentration of OH^- we are driving the value of E to zero. If our reduction half-reaction is less "willing" to accept electrons then our oxidation half-reaction is less able to donate them. The Nernst equation and the principles behind it combine the concepts of electrochemistry and equilibrium.

Like I said though, I'm also a student, so take what I say with a grain of NaCl.