Explaining The Non-Reaction Of An Iron Nail With Equilibrium

[Logan Johnston]

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)

Originally I had the thought to regard the electron concentration as an indicator of equilibrium, thinking that since the electron concentration is shifted reactant side, no reaction will occur. Now I'm beginning to second guess that hypothesis because the half reaction doesn't explicitly deal with the production of an iron oxide, therefore I find it a presumptuous to say the electron concentration in the OH-(aq) half reaction dictates a successful redox reaction.

Any thoughts?

 

[Logan Johnston]

To Be Clear: The reaction of iron in NaOH is spontaneous, I know. If I could change the title I would.

 

[epenguin]

Yeah I had a thought, but only a simple one.
I thought about mass action.
I'm not capable of having a thought about electron concentration - I never had any lessons that mentioned it, did you?
(You are clear though where these e^- are coming from though?)
Not happy how complete or correct my explanation is but it is at least rational.
Corrosion is quite complicated with stuff you could not be expected to all deduce from first principles - but those principles are useful for understanding the explanations. I have not found anything outstandingly good on it on the web, particularly not on pH effect. Wiki reasonable. However an awareness of the main methods, particularly electrochemical, of rust prevention is probably a reasonable (exam question) expectation at your level.

 

[Borek]

Basic idea is that we can express Nernst equation as

E = E_0 + \frac {RT}{nF} ln \frac {[Ox]}{[Red]} = \frac {RT}{nF} ln \frac {[Ox][e^-]}{[Red]}

where

[e^-] = e^{\frac{nFE_0}{RT}}

Just a mathematical trick, but sometimes useful.

 

[Logan Johnston]

epenguin: Iron giveth and hydroxide taketh thee electrons so to speak, yes. I've not yet had any lessons concerning electron concentration that could apply, no, it was more of an intuitive thought. Certainly from an electrochemical perspective I can describe the redox of iron in NaOH, but from an equilibrium perspective I must be missing something. I like the thought of mass action because it deals with equilibrium, but I will have a reread of my chapters tonight and see if I read over something.

Borek: That's very interesting. I'll have a study of that tonight.

Thank you both, I'll see what I can do later today.

 

[Logan Johnston]

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 all for your input.

 

[epenguin]

if we add more hydroxide ions, the reaction will shift towards the left, correct? This will impede the reduction half-reaction
.

That is exactly what I meant by

I thought about mass action.

I am glad To see you came back though - too many threads here are not concluded, although the students probably have answers as you have. "when you have an announcer it's not finished" - there is often a little something that it is useful to add, not at all answers are as good as the students imagine

 

[Logan Johnston]

I thought about mass action.

I understand that now, absolutely.

After a bit more thinking (and watching a diploma review video) I realized that the given half reaction is necessary for all rusting examples in this level of chemistry; something I had not been explicitly taught. So when that half-reaction is applied to the scenario of an iron nail in hydroxide solution, the increased concentration of hydroxide favours the reverse reaction and no rusting occurs.

If anyone with the same question as I had looks at this in the future, I was wrong to say that iron reacts spontaneously with NaOH solution. The law of mass action dictates the reverse reaction be favoured in order to restore equilibrium to the system, so the reaction is not spontaneous.