Balancing complex redox reaction of gold treated with sodium cyanide

Question

How would you balance this reaction:

$$\ce{Au(s) + NaCN(aq) + O2(g) + H2O(l)-> Na[Au(CN)2](aq) + NaOH(aq)}?$$

Here’s what I’m came up with. First we can balance any atoms besides hydrogen and oxygen in the reaction (like $\ce{Na}$ and $\ce{CN}):$

$$\ce{Au(s) + 2 NaCN(aq) + O2(g) + H2O(l) -> Na[Au(CN)2](aq) + NaOH(aq)}$$

Then we can write out the half reactions:

$$ \begin{align} \text{Anode:} &\quad &\ce{Au(s) &-> Au+} \\ \text{Cathode:} &\quad &\ce{O2(g) &-> H2O} \end{align} $$

Since the oxygen is going from an oxidation state of $0$ to $-2$ in the cathode half reaction, I figured you can rewrite $\ce{OH-}$ as $\ce{H2O}$ on the right hand side of the equation.

For the half reactions, I figured we can eliminate the ions whose oxidation number don’t change but I’m not sure whether you should include $\ce{NaOH}$ in the anode or cathode half reaction.

Answer 1

In principle, the oxidation of gold is the half-reaction $$\ce{Au -> Au^+ + e^-}$$But as this oxidation is done in a cyanide medium, $\ce{Au^+}$ is included in a complex ion : $$\ce{Au + 2 CN^- -> Au(CN)_2^- + e^-}$$ The second half-equation is $$\ce{2H2O + O2 + 4 e^- -> 4 OH^-}$$ So the final equation describing the dissolution of gold in an aerated cyanide solution is the sum of the two previous equations, after multiplying the first by 4 : $$\ce{4 Au + 8 CN^- + 2 H2O + O2 -> 4 Au(CN)2^- + 4 OH^-}$$ And if you prefer equations with neutral substances, you may write it this way $$\ce{4 Au + 8 NaCN + 2 H2O + O2 -> 4 NaAu(CN)2 + 4 NaOH}$$