Question

When an aqueous solution containing gold(III) salt is electrolyzed, metallic gold is deposited at the cathode and oxygen gas is generated at the anode. (a) If \(9.26 \mathrm{~g}\) of Au is deposited at the cathode, calculate the volume (in liters) of \(\mathrm{O}_{2}\) generated at \(23^{\circ} \mathrm{C}\) and \(747 \mathrm{mmHg}\). (b) What is the current used if the electrolytic process took \(2.00 \mathrm{~h} ?\)

Short answer

(a) Volume of \(\mathrm{O}_2\) is approximately 0.872 L. (b) The current used is approximately 1.89 A.

Step-by-step solution

Write the relevant half-reactions

For the electrolytic process, the reduction reaction at the cathode is: \[ \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \]. At the anode, the oxidation reaction is: \[ 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \].

Calculate moles of gold deposited

Given mass of gold (Au) deposited is \(9.26\, \mathrm{g}\). The molar mass of gold is \(197.0\, \mathrm{g/mol}\). Calculate the moles of Au deposited: \[ \text{moles of Au} = \frac{9.26\, \mathrm{g}}{197.0\, \mathrm{g/mol}} = 0.0470\, \mathrm{mol} \].

Determine moles of electrons transferred

From the reduction reaction \(\text{Au}^{3+} + 3\text{e}^- \rightarrow \text{Au}\), 3 moles of electrons are needed to deposit 1 mole of Au. Thus, moles of electrons: \[ 0.0470\, \text{mol Au} \times 3 = 0.141\, \text{mol e}^- \].

Calculate moles of oxygen gas produced

From the oxidation reaction \(2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-\), 4 moles of electrons produce 1 mole of \(\text{O}_2\). Calculate moles of \(\text{O}_2\): \[ \frac{0.141\, \text{mol e}^-}{4} = 0.0353\, \text{mol O}_2 \].

Calculate volume of oxygen gas at given conditions

Use the ideal gas law \(P V = nRT\). Convert pressure: \(747\, \text{mmHg} = 747 / 760\, \text{atm} = 0.983\, \text{atm}\). Convert temperature to Kelvin: \(T = 23 + 273.15 = 296.15\, \text{K}\). Using \(R = 0.0821\, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}\), solve for \(V\): \[ V = \frac{nRT}{P} = \frac{0.0353 \times 0.0821 \times 296.15}{0.983} \approx 0.872\, \text{L} \].

Calculate current used in electrolysis

Use Faraday's law: \(Q = nF\), where \(n\) is the moles of electrons, and \(F = 96485\, \text{C/mol}\) is Faraday's constant. Calculate charge: \[ Q = 0.141\, \text{mol e}^- \times 96485\, \text{C/mol} = 13600\, \text{C} \]. Current \(I\) is \(I = \frac{Q}{t}\). Convert time to seconds: \(2\, \text{hours} = 7200\, \text{seconds}\). Solve for \(I\): \[ I = \frac{13600\, \text{C}}{7200\, \text{s}} \approx 1.89\, \text{A} \].

Key concepts

Oxidation and Reduction Reactions

In electrochemistry, oxidation and reduction reactions are key processes that involve the transfer of electrons. These reactions can be represented as half-reactions that occur at the anode and the cathode in the electrochemical cell. In the given exercise, at the cathode, we have a reduction reaction:

• Gold ions \(( \text{Au}^{3+} )\) gain electrons and are reduced to metallic gold \( \text{Au} \).

On the other hand, at the anode, oxidation occurs:

• Water molecules are oxidized to produce oxygen gas \( \text{O}_2 \), hydrogen ions, and electrons.

Understanding the flow of electrons in these half-reactions helps in calculating other parameters of the electrolysis process, such as the number of moles of substances produced or consumed. This is crucial for determining industrial applications, such as metal extraction and purification.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that connects the pressure, volume, temperature, and amount of gas present within a system. It is expressed as:\[ PV = nRT \]where:

• \( P \) is the pressure,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature in Kelvin.

In our exercise, the Ideal Gas Law is used to find the volume of oxygen gas produced during electrolysis. By converting the conditions to the required units (atm for pressure, Kelvin for temperature), we can rearrange the equation to solve for volume \( V \). The application of this law helps in obtaining the volume of gases under various conditions in both academic and industrial settings.

Faraday's Law

Faraday's Law of Electrolysis provides a quantitative relationship between the amount of substance reacted at each electrode during electrolysis and the quantity of electric charge (Q) passed through the circuit. Faraday's Law can be stated as:\[ Q = nF \]where:

• \( Q \) is the total electric charge,
• \( n \) is the number of moles of electrons transferred, and
• \( F \) is Faraday's constant (\(96485\, \text{C/mol}\)).

In our problem, we use this relationship to determine the charge required to deposit a certain amount of gold and, subsequently, to calculate the current used in the process. This theoretical framework is particularly critical in applications like electroplating, electrorefining, and electro-synthesis, where precise control over the deposited materials is needed.

Electrolysis

Electrolysis is a chemical process that uses electrical current to drive a non-spontaneous chemical reaction. In the given exercise, an aqueous solution containing gold(III) salt undergoes electrolysis, resulting in metallic gold being deposited at the cathode and oxygen gas being generated at the anode.

• The process involves passing a direct current through the solution, causing the chemical changes at both electrodes based on the provided half-reactions.
• This method is used efficiently for electroplating metals, refining ores, and producing elements from their compounds.

Understanding electrolysis helps in grasping how energy conversion is used to achieve chemical transformations that would not naturally occur, emphasizing its significance in both industrial processes and research applications.

Gas Volume Calculation

Calculating the volume of a gas produced or consumed in a chemical reaction is essential, especially in processes like electrolysis where gases like oxygen are generated. In the exercise, we calculate the volume of oxygen gas produced using the moles of gas determined from its relation to electron transfer.

• The Ideal Gas Law comes into play to convert moles of gas into volume under specified temperature and pressure conditions.
• Converting pressure into atmospheres and temperature into Kelvin is important to ensure that the units are compatible with the gas constant \( R \).

By accurately calculating gas volumes, you can better understand reaction stoichiometry and design systems for efficient industrial gas collection and utilization. These calculations are vital in fields like environmental science, chemical engineering, and materials science, where gas production is a significant aspect of various processes.