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

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

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

$$\ce{Fe + CuSO4 -&gt;~ FeSO4 + Cu}$$