How do I determine whether it is the *ferrous* or *ferric* ion that is implied in this displacement reaction?

I'm supposed to write a balanced equation for:

$$\ce{Iron(s) + copper(II) sulfate(aq) -> ?}$$

If I use Iron(II)/ferrous I get

$$\ce{Fe + CuSO4 -> FeSO4 + Cu}$$

and if I use Iron(III)/ferric,

$$\ce{Fe + CuSO4 -> Fe2(SO4)3 + Cu}$$

(Before balancing).

Is there a way to determine which one I am supposed to use?

• Tip: Fe(III) salts are used in etching of copper for making custom electric boards by enthusiasts. – permeakra Jun 12 '14 at 18:07
• Yes, it is. Generally, you should split the reactions into half-reactions and then consult standard electrode potential table. In case differences are large, results are obvious. In case differences are small (<0.2-0.5 V), some calculations involving Nernst equation must be performed. This, of course, have limited use, since kinetic differences are not considered, but for metal substitution for salts such procedures usually works. For details consult electrochemistry chapter of your favorite physical chemistry textbook. I'm too lazy to write a proper answer here. – permeakra Jun 12 '14 at 18:14
• @permeakra it's good you are not at least not lazy in writing answer in comment. I am even lazy in that. That's why i normally give some useful links :) – Freddy Jun 13 '14 at 7:18

You can determine which reaction takes place by the standard electrode potentials of the half reactions. The relevant standard reduction potentials for the possible reactions (source of the values) are:

$$\ce{Cu^2+ +2e- ->~ Cu} \quad E_0=+0.342\mathrm{V}$$ $$\ce{Fe^2+ + 2e- ->~ Fe} \quad E_0=-0.447\mathrm{V}$$ $$\ce{Fe^3+ + e- ->~ Fe^{2+}} \quad E_0=+0.771\mathrm{V}$$

The overall potential of the electrochemical cell is defined as the difference between the standard reduction potentials of reduction and oxidation reaction.

$$E_{cell}=E_{red}-E_{ox}$$

For a spontaneous reaction to occur, $E_{cell}$ must be positive. With the potentials of copper reduction and oxidation of iron to $\ce{Fe^2+}$, you get

$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^2+/Fe}}=+0.342\mathrm{V}-(-0.447\mathrm{V})=+0.789\mathrm{V}$$

It is obvious from the values that $\ce{Cu^2+}$ cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ because $E_{cell}$ is negative for this reaction.

$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^3+/Fe^2+}}=+0.342\mathrm{V}-(+0.771\mathrm{V})=-0.429\mathrm{V}$$

The balanced equation for the redox reaction between iron and copper(II) sulfate is therefore

$$\ce{Fe + CuSO4 ->~ FeSO4 + Cu}$$