Is it possible to balance this equation using the half reaction method?

$$\ce{Au + Cl2 + HCl⟶HAuCl4}$$

I know that the equation balanced would be: $$\ce{2Au + 3Cl2 + 2HCl⟶2HAuCl4}$$

However, how would the half reactions end up?


1 Answer 1


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:

$$\ce{Au -> [Au^3+] + 3 e-}\tag{Ox}$$

$$\ce{Cl2 + 2 e- -> 2 Cl-}\tag{Red}$$

Which gives you:

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

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

$$\ce{AuCl3 + HCl -> H[AuCl4]}\tag{Aurate}$$

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.


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