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I plan on conducting an experiment in which I place an iron nail in water and analyze its rate of corrosion into rust with different concentrations of NaCl.

My original plan was to cover the water in a beaker with oil then use a dissolved oxygen probe to measure the drop in oxygen levels as the reaction went on. However, I was thinking that maybe conductivity would change as the reaction went on?

In short, in a solution containing iron, salt, oxygen, and water, will conductivity change as the iron rusts?

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This is an interesting question. As iron rusts, you will have a couple of reactions. First, as you expect is the oxidation of iron:

$\ce{Fe ->Fe^2+ + 2e^-}$ or $\ce{Fe ->Fe^3+ + 3e^-}$

Then, either one of these may precipitate as their oxide, or hydroxide or their chloride.

On the flip side, somewhere else on the same piece of nail, there has to be a reduction. This may be one of:

$\ce {2H+ +2e- -> H2} $ or $\ce{2O2 + 4e- +4H+ ->2H2O}$

The overall reaction will probably be a combination of the above depending on the various rates of reaction. A priori, I would say it is very hard to predict which mechanism would be operative and therefore it is hard to predict which way conductivity would go.

Another point is even if the overall mechanism was such that there was some imbalance that would affect the conductivity, that would be very hard to measure. The changes would be a very small fraction of the total conductivity which would make it a hard measurement.

The typical way this is measured is "weight loss", which involves cleaning the corrosion products off the metal specimen and measuring the weight after given time periods. You can find lots of cleaning procedures and details of weight loss measurements over the net.

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    $\begingroup$ Assuming starting out with distilled water, both reduction half cell reactions consume $\ce{H+}$. Thus the solution would get more basic. $\ce{O2}$ is uncharged and therefore not conductive. Since $\ce{H+}$ has a greater conductivity than $\ce{OH-}$, the conductivity of the solution must go down. $\endgroup$ – MaxW Jan 28 '17 at 22:03

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