Identify the oxidation and reduction half-reactions. In this case, Au3+ is reduced to Au, and Cr3+ is oxidized to Cr2O7^2-.
Write the reduction half-reaction: Au3+(aq) + 3 e^- → Au(s). This shows the gain of electrons by gold ions.
Write the oxidation half-reaction: 2 Cr3+(aq) + 7 H2O(l) → Cr2O7^2-(aq) + 14 H+(aq) + 6 e^-. This shows the loss of electrons by chromium ions.
Balance the electrons between the two half-reactions. Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1 to balance the electrons exchanged.
Combine the balanced half-reactions to form the overall balanced equation: 6 Au3+(aq) + 2 Cr3+(aq) + 7 H2O(l) → 6 Au(s) + Cr2O7^2-(aq) + 14 H+(aq).