Question

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 \(1.00 \mathrm{~mol}\) gold?

Short answer

During the electrochemical production of 1.00 mol of gold, 0.50 mol of oxygen gas will be produced.

Step-by-step solution

Half-reactions

Write down the half-reactions occurring at the electrodes for this electrochemical process. At the cathode (reduction): Au(CN)₂⁻ + e⁻ → Au + 2CN⁻ At the anode (oxidation): H₂O → 2H⁺ + 1/2O₂ + 2e⁻

Balance the half-reactions

Multiply the half-reactions by the appropriate factors so that the number of electrons lost and gained are the same (the electron transfer is conserved). For the reduction half-reaction: 2(Au(CN)₂⁻ + e⁻ → Au + 2CN⁻) For the oxidation half-reaction: 4(H₂O → 2H⁺ + 1/2O₂ + 2e⁻) Now, the number of electrons transferred is balanced.

Write the balanced overall reaction

Combine the balanced half-reactions to write the balanced overall reaction. 2Au(CN)₂⁻ + 4H₂O → 2Au + 4H⁺ + 2O₂ + 4CN⁻

Calculate the amount of O₂ produced per mol of gold

Utilize stoichiometry to determine the amount of O₂ that will be produced along with 1.00 mol of gold. According to the balanced overall reaction, 2 moles of gold are produced along with 2 moles of O₂. Thus, for the production of 1 mole of gold, we can determine the moles of O₂ formed: \(1.00 \thinspace \text{mol Au} \times \frac{1 \thinspace\text{mol O}_2}{2 \thinspace\text{mol Au}} = 0.50 \thinspace\text{mol O}_2\)

Conclusion

During the electrochemical production of 1.00 mol of gold, 0.50 mol of oxygen gas will be produced.

Key concepts

Half-reactions

In electrochemical processes, reactions are split into two halves: a reduction half-reaction and an oxidation half-reaction. This division simplifies balancing elements and electrons.
At the cathode, where reduction happens, ions or molecules gain electrons. For our gold process, the half-reaction is:
- \[\mathrm{Au(CN)}_2^- + \mathrm{e}^- \to \mathrm{Au} + 2\mathrm{CN}^- \]
Anions in solution (\(\mathrm{Au(CN)}_2^- \)) receive electrons to form solid \(\mathrm{Au}\).

On the other hand, the anode facilitates oxidation, where molecules lose electrons. For this exercise, water molecules are oxidized:
- \[\mathrm{H}_2\mathrm{O} \to 2\mathrm{H}^+ + \frac{1}{2}\mathrm{O}_2 + 2\mathrm{e}^-\]
Oxidation releases electrons and forms \(\mathrm{O}_2\), illustrating the elemental change crucial to electrochemical processes.

Stoichiometry

Stoichiometry is the mathematical approach in chemistry that involves the conversion between moles of different substances based on a balanced chemical equation. It's pivotal for quantifying reactions.
In the provided exercise, stoichiometry is used to determine how much oxygen (\(\mathrm{O}_2\)) is produced with 1 mol of gold (Au). According to the balanced equation, 2 moles of \(\mathrm{Au}\) will produce 2 moles of \(\mathrm{O}_2\).

• Therefore, for the production of 1 mol of Au, \(0.50\) mol of \(\mathrm{O}_2\) is formed since the reaction's stoichiometry shows a 1:1 ratio after the initial factors are considered.

Understanding stoichiometry is essential as it allows chemists to predict product formation and reactant requirements, ensuring reaction efficiency.

Electrolysis

Electrolysis is the technique of using direct electric current to drive a non-spontaneous chemical reaction. It's a central process in the electrochemical production of elements like gold.
This method involves the breakdown of ionic compounds. Here, during electrolysis, the decomposition of \(\mathrm{Au(CN)}_2^-\) occurs at the cathode:

• Reduction: Electrons are supplied to the dissolved species, converting them to solid gold.
• Oxidation: This accompanies the reduction at the anode where water releases \(\mathrm{O}_2\) and electrons.

Electrolysis is crucial in industries for extracting pure forms of reactive elements from their compounds, enabling large-scale production of metals like gold with high purity.

Balancing Chemical Equations

Balancing chemical equations ensures the conservation of mass and charge - a fundamental chemistry principle. For any reaction, the number of atoms for each element becomes equal on both sides of the equation.
Balancing involves:

• Identifying each reactant and product in the reaction.
• Ensuring equal numbers of atoms by adjusting coefficients before the formulas.

In this exercise, the balanced equation serves as the ultimate goal for understanding the chemical process:
\[2\mathrm{Au(CN)}_2^- + 4\mathrm{H}_2\mathrm{O} \to 2\mathrm{Au} + 4\mathrm{H}^+ + 2\mathrm{O}_2 + 4\mathrm{CN}^-\]
Each part of this reaction is meticulously adjusted to reflect accurate stoichiometric relationships. This accuracy is necessary for reliable predictions in reactions and computations involving mass and energy transfers.