Balancing complex redox reaction of gold treated with sodium cyanide How would you balance this reaction:
$$\ce{Au(s) + NaCN(aq) + O2(g) + H2O(l)-&gt; Na[Au(CN)2](aq) + NaOH(aq)}?$$
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 $\ce{CN}):$
$$\ce{Au(s) + 2 NaCN(aq) + O2(g) + H2O(l) -&gt; Na[Au(CN)2](aq) + NaOH(aq)}$$
Then we can write out the half reactions:
$$
\begin{align}
\text{Anode:} &amp;\quad &amp;\ce{Au(s) &amp;-&gt; Au+} \\
\text{Cathode:} &amp;\quad &amp;\ce{O2(g) &amp;-&gt; H2O}
\end{align}
$$
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.
In principle, the oxidation of gold is the half-reaction $$\ce{Au -&gt;  Au^+ + e^-}$$But as this oxidation is done in a cyanide medium, $\ce{Au^+}$ is included in a complex ion : $$\ce{Au  +  2 CN^- -&gt; Au(CN)_2^- + e^-}$$ The second half-equation is $$\ce{2H2O + O2 + 4 e^- -&gt; 4 OH^-}$$ 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 : $$\ce{4 Au + 8 CN^- + 2 H2O + O2 -&gt; 4 Au(CN)2^- + 4 OH^-}$$ And if you prefer equations with neutral substances, you may write it this way $$\ce{4 Au + 8 NaCN + 2 H2O + O2 -&gt; 4 NaAu(CN)2 + 4 NaOH}$$