Step 1: Recall the Daniell cell setup.
A Daniell cell consists of two half-cells: one containing zinc metal and zinc sulfate (\(Zn | Zn^{2+}\)) and the other containing copper metal and copper sulfate (\(Cu | Cu^{2+}\)). The zinc electrode is the anode, where oxidation takes place, and the copper electrode is the cathode, where reduction occurs.
Step 2: Electrons and current flow.
In the Daniell cell, oxidation occurs at the zinc electrode:
\[
Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-
\]
The electrons released by zinc flow through the external circuit towards the copper electrode. Therefore, the electrons flow from zinc electrode to copper electrode, which means current flows in the opposite direction, from copper to zinc. So, option \((2)\) is correct.
Step 3: Cation movement.
In the electrolyte solution, cations like \(Zn^{2+}\) move towards the cathode (copper electrode), and anions move towards the anode (zinc electrode). Thus, the cations move toward the copper electrode.
Step 4: Reduction at cathode.
At the cathode, reduction occurs:
\[
Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)
\]
Thus, reduction occurs at the copper electrode, which is the cathode.
Step 5: Conclusion.
Thus, the correct statement is option \((2)\), where current flows from the zinc electrode to the copper electrode.
Step 6: Final answer.
\[
\boxed{\text{Current flows from zinc electrode to copper electrode.}}
\]