Gold is produced electrochemically from an aqueous solution of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) containing an excess of \(\mathrm{CN}^{-}\). Gold metal and oxygen gas are produced at the electrodes. What amount (moles) of \(\mathrm{O}_{2}\) will be produced during the production of \(1.00 \mathrm{~mole}\) of gold?

First, let's write down the half-reactions for this electrochemical process: Oxidation half-reaction (at the anode): \[\mathrm{OH}^{-} \rightarrow \frac{1}{2}\mathrm{O}_{2} + \mathrm{H}_{2}\mathrm{O} + 2\mathrm{e}^{-}\] Reduction half-reaction (at the cathode): \[\mathrm{Au}(\mathrm{CN})_{2}^{-} + \mathrm{e}^{-} \rightarrow \mathrm{Au} + 2\mathrm{CN}^{-}\]

To balance the half-reactions, we need to make sure that the same number of electrons are being gained and lost in both of the half-reactions. Since there are already 2 electrons lost in the oxidation half-reaction and 1 electron gained in the reduction half-reaction, we can balance the reduction half-reaction by multiplying it by 2. Oxidation half-reaction: \[\mathrm{OH}^{-} \rightarrow \frac{1}{2} \mathrm{O}_{2} + \mathrm{H}_{2}\mathrm{O} + 2\mathrm{e}^{-}\] Balanced reduction half-reaction: \[2\mathrm{Au}(\mathrm{CN})_{2}^{-} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Au} + 4\mathrm{CN}^{-}\]

Now that we have balanced the half-reactions, we can combine them to get the overall balanced chemical equation for the process. \[\mathrm{OH}^{-} + 2\mathrm{Au}(\mathrm{CN})_{2}^{-} \rightarrow \frac{1}{2} \mathrm{O}_{2} + \mathrm{H}_{2}\mathrm{O} + 2\mathrm{Au} + 4\mathrm{CN}^{-}\]

Now that we have the overall balanced chemical equation, we can use stoichiometry to find the amount of oxygen gas produced during the production of 1.00 mole of gold. From the balanced equation, we can see that for every 2 moles of gold formed, 0.5 moles of oxygen gas is produced. Therefore, for 1.00 mole of gold formed, we can calculate the amount of oxygen gas produced as follows: \[\frac{\text{moles of}\ \mathrm{O}_{2}}{\text{moles of}\ \mathrm{Au}} = \frac{1}{4}\] \[ \text{moles of}\ \mathrm{O}_{2} = \frac{1}{4} \times \text{moles of}\ \mathrm{Au} \] \[ \text{moles of}\ \mathrm{O}_{2} = \frac{1}{4} \times 1.00 \] \[ \text{moles of}\ \mathrm{O}_{2} = 0.25\] So, 0.25 moles of oxygen gas will be produced during the production of 1.00 mole of gold.