Balancing complex redox reaction of gold treated with sodium cyanide
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 $$\ce{4 Au + 8 NaCN + 2 H2O + O2 -&gt; 4 NaAu(CN)2 + 4 NaOH}$$