Balancing a redox equation with only one product
You actually have different possibilities of setting up half-reactions. One way is to note that formally, gold goes from a $\pm 0$ to a $\mathrm{+III}$ oxidation state, thus it can be written as:

$$\ce{Au -&gt; [Au^3+] + 3 e-}\tag{Ox}$$

$$\ce{Cl2 + 2 e- -&gt; 2 Cl-}\tag{Red}$$

Which gives you:

$$\ce{2 Au + 3 Cl2 -&gt; 2 [Au^3+] + 6 Cl- -&gt; 2 AuCl3}\tag{Redox}$$

And, if additional $\ce{HCl}$ is present:

$$\ce{AuCl3 + HCl -&gt; H[AuCl4]}\tag{Aurate}$$

Since $\ce{HCl}$ does not participate in the redox part of the reaction, it does not occur until after the redox part.

Naturally, $\ce{[Au^3+]}$ is not isolateable under these conditions, hence the square brackets.