Using these equations: \begin{aligned} \text{(Fe2):}&&\ce{ Fe + CuSO4 &-> FeSO4 + Cu}\\ \text{(Fe3):}&&\ce{ 2Fe + 3CuSO4 &-> Fe2(SO4)3 + 3Cu} \end{aligned}

Beginning with $\pu{0.78 g}$ of iron, the theoretical yields of copper are $\pu{0.89 g}$ (with $\ce{Fe^{2+}}$) and $\pu{1.33 g}$ (with $\ce{Fe^{3+}}$), and the actual yield was $\pu{0.78 g}$ of copper. I know to find of the ratio of the moles of iron used to the moles of copper formed (88% for $\ce{Fe^{2+}}$, 58% for $\ce{Fe^{3+}}$), and so the clear choice would be that this reaction formed $\ce{Fe^{2+}}$ ions.

But my teacher said this would form $\ce{Fe^{3+}}$ ions, but why is that? Shouldn't it have formed $\ce{Fe^{2+}}$? The reaction formed $\pu{0.12 mol}$ of copper, compared to the $\pu{0.14 mol}$ in the theoretical yield, and the other $\pu{0.02 mol}$ could have just been lost somewhere along the reaction process.


When "doing" science, usually you try to (in-)validate a model of the workings of the world. In the case of chemists, this world is mostly confined to the scale of molecules.

That said, let's proceed with

A Theoretical Model

Listed below are some standard reduction potentials I've gathered from the Handbook of Chemistry and Physics, 95th ed. (probably paywalled for you): $$ \begin{align} &&E^0_\text{red} / \mathrm{V}\\ \hline \ce{Fe^{2+} + 2e- & <=> Fe} & -0.447 \\ \ce{Fe^{3+} + 3e- & <=> Fe} & -0.037 \\ \ce{Cu^{2+} + 2e- & <=> Cu} & 0.342 \\ \end{align} $$

Now we can calculate the potential (and the resulting electromotive force EMF) for each of the reactions: $$ \begin{align} E^0_\text{(Fe2)} = E^{0}_\text{red}(\text{red}) - E^0_\text{red}(\text{ox}) &= 0.342~\mathrm{V} - (-0.447~\mathrm{V}) = 0.789~\mathrm{V} \\ E^0_\text{(Fe3)} &= 0.342~\mathrm{V} - (-0.037~\mathrm{V}) = 0.379~\mathrm{V} \end{align} $$

It is clear that the reaction (Fe2) releases more energy and thus should be preferred over the other reaction. This means that we expect a yield of 100% Cu and not 150% Cu.

Now that we have gathered some theoretical understanding of the matter, let's proceed to the interesting part, which is

Confirmation With Experimental Results

You have correctly calculated the theoretical yields for copper to be 0.89 g and 1.33 g, respectively. The corresponding amounts of copper are 0.014 mol and 0.021 mol.

You have measured the amount of copper produced, and it clocks in at 0.78 g. This is already a strong indication that we should side with (Fe2) on this one. But just to really be sure, let's look at the error that we would have if it was (Fe3): $$ \varepsilon = \frac{\Delta x}{x} = 41.4\% $$ This is a very high error, especially when compared to the error for (Fe2), which is only 12.1%. Assuming that you did not perform the reaction with pure Fe but a mixture of iron with some iron oxides as contaminants, did not use an analytical scale and performed the experiment in fairly uncontrolled conditions (are you sure the reaction was complete when you aborted?) this error is not as bad as it seems.

With the experimental results in our pocket, we can now move on to the final part:

The Conclusion

We have shown, using theoretical predictions and experimental confirmation, that the reaction observed was the oxidation of iron to iron(II) under reduction of copper(II) to copper.

By the way, most of the calculations I did with an iPython Notebook, which you can view here. I might have messed something up, you never know.

  • 1
    $\begingroup$ Your analysis misses a subtle point. If you consider just the ocidationnof Fe(II) to Fe(III), that has a standard potential of +0.783 V. Therefore Fe(II) is more stable at the potential required to reduce copper, and in fact Fe(III) would oxidize copper metal back to Cu(II) (which then reacts with more elemental iron). This reaction cycle tends to have ugly consequences in pickling of steel strip. $\endgroup$ Sep 15 '20 at 16:23

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