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Redox reactions are a fundamental concept in electrochemistry, involving the transfer of electrons between two chemical species. This process is split into oxidation and reduction reactions.
During oxidation, a substance loses electrons, while reduction involves the gain of electrons. In our given problem, to produce gold metal and oxygen gas, we have separate reactions occurring at the electrodes.
At the anode, oxidation takes place where cyanide ions (\(\mathrm{CN}^{-}\)) lose electrons, leading to the formation of oxygen gas (\(\mathrm{O}_{2}\)). At the cathode, gold ions are reduced as they gain electrons to form solid gold (\(\mathrm{Au}\)).

• The anode reaction: \(\mathrm{CN}^{-}(aq) \rightarrow \frac{1}{2}\mathrm{O}_{2}(g) + \mathrm{CNO}^{-}(aq) + e^{-}\).
• The cathode reaction: \(\mathrm{Au}(\mathrm{CN})_{2}^{-}(aq) + e^{-} \rightarrow \mathrm{Au}(s) + 2\mathrm{CN}^{-}(aq)\).

Understanding these reactions is key to comprehending how the overall electrolytic process separates and forms specific substances at each electrode. This separation is the driving force behind electrochemical processes, allowing valuable materials such as gold to be extracted and refined efficiently.

Stoichiometry is an essential concept in chemistry, which involves measuring and calculating the quantities of reactants and products in chemical reactions.
It rests on the idea that elements are conserved in chemical reactions and allows us to determine the proportions in which substances react.
In our problem, stoichiometry is used to find out how much oxygen gas is produced when a specific amount of gold is formed.
The balanced chemical equation reveals that for every 2 moles of gold (\(\mathrm{Au}\)), 1 mole of oxygen gas (\(\mathrm{O}_{2}\)) is generated. To solve the problem, we start with the amount of gold produced and use the stoichiometric ratio to calculate the corresponding oxygen amount:

• The ratio calculated from the balanced equation is \(\frac{1\ \text{mole of}\ \mathrm{O}_2}{2\ \text{moles of}\ \mathrm{Au}}\).
• Thus, when 1 mole of gold is produced, \(0.5\ \text{mole of}\ \mathrm{O}_2\) is generated.

this simple but powerful relationship indicates how stoichiometric calculations provide quantitative insights into chemical reactions, ensuring that all reactants are accounted for and used efficiently.

Electrode reactions are critical components of electrochemical processes, where oxidation and reduction occur at different sites, namely the anode and the cathode. These reactions dictate the overall reaction happening in an electrochemical cell.
Each electrode is involved in a half-reaction, with specific processes taking place:

• At the anode, oxidation occurs, which in this problem involves the transformation of cyanide ions to oxygen gas and cyanate ions.
• At the cathode, reduction is observed, with gold ions gaining electrons to form solid gold.

The half-reactions need to be balanced, not just in elements but also in charge to ensure that the overall electrical neutrality is maintained within the cell.
In the given exercise, the oxidation at the anode and the reduction at the cathode combine to produce oxygen gas and metallic gold while maintaining a continuous flow of electrons through the cell.
Understanding these reactions is essential for predicting the products of electrochemical processes and is foundational for industrial applications like electroplating and metal refining.