Use the galvanic cell reaction to answer the question.

In a galvanic cell, the cathode is where reduction occurs, meaning that it is the site where electrons are gained.

In the given reaction:

Oxidation half-reaction (at the anode): \[ 2Cr(s) \rightarrow 2Cr^{3+}(aq) + 6e^- \]

Reduction half-reaction (at the cathode): \[ 3Cu^{2+}(aq) + 6e^- \rightarrow 3Cu(s) \]

Thus, the half-reaction that occurs at the cathode is:

\[ 3Cu^{2+}(aq) + 6e^- \rightarrow 3Cu(s) \]

This indicates that copper ions (\(Cu^{2+}\)) are being reduced to solid copper (\(Cu\)).