Problem 73 - Chemistry: An Atoms-Focused Approach

Question

A quantity of electric charge deposits \(0.732 \mathrm{g}\) of \(\mathrm{Ag}(\mathrm{s})\) from an aqueous solution of silver nitrate. When that same quantity of charge is passed through a solution of a gold salt, \(0.446 \mathrm{g}\) of \(\mathrm{Au}(s)\) is formed. What is the oxidation state of the gold ion in the salt?

Step-by-Step Solution

Answer

Answer: The oxidation state of the gold ions in the gold salt is +3.

Step 1: Determine the moles of silver and gold deposited

The given mass of silver and gold can be used to calculate their moles using their respective molar masses. Note that the molar mass of silver (Ag) is 107.87 g/mol and that of gold (Au) is 197.0 g/mol. For silver: \(\text{moles of Ag} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.732 g}{107.87 g/mol} = 6.78\times 10^{-3}\,\text{mol}\) For gold: \(\text{moles of Au} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.446 g}{197.0 g/mol} = 2.26\times 10^{-3}\,\text{mol}\)

Step 2: Write the chemical equation for the reduction of silver ions

The chemical equation for the reduction of silver ions in an aqueous solution of silver nitrate describes the formation of silver atoms when a certain amount of electric charge passes through the solution. The equation for the reduction of silver ions is \(\text{Ag}^+\,(aq) + e^- \rightarrow \text{Ag}\,(s)\) From the stoichiometry of this reaction, we can see that 1 mole of silver ions requires 1 mole of electrons to be reduced to 1 mole of silver atoms.

Step 3: Calculate the moles of electrons transferred in the deposition of silver

Using the stoichiometry of the reduction of silver ions from step 2 and the calculated moles of silver from step 1, we can find the moles of electrons transferred in the deposition of silver. \(\text{moles of electrons} = \text{moles of Ag} = 6.78\times 10^{-3}\,\text{mol}\)

Step 4: Write the chemical equation for the reduction of gold ions

Using the moles of electrons transferred in the deposition of silver to form gold atoms, we will determine the reduction equation for gold ions and the number of electrons required for its reduction. Let "x" be the oxidation state of gold ion and "e" the electrons required. \(\text{Au}^{x+}\,(aq) + x\,e^- \rightarrow \text{Au}\,(s)\)

Step 5: Calculate the oxidation state of gold ion

From step 3, we have the moles of transferred electrons. Since the same electric charge deposits both silver and gold, these moles of electrons are also involved in the deposition of gold. Using the stoichiometry of the gold reduction equation from step 4 and the calculated moles of gold from step 1, we can determine the oxidation state (x) of gold ion. \(\text{moles of electrons} = x \times \text{moles of Au}\) Solving for x: \(x = \frac{\text{moles of electrons}}{\text{moles of Au}} = \frac{6.78\times 10^{-3}\,\text{mol}}{2.26\times 10^{-3}\,\text{mol}} \approx 3\) The oxidation state of the gold ions in the gold salt is +3.

Key Concepts

Electrochemistry

Electrochemistry involves the study of chemical reactions that produce electrical currents and vice versa. At its core, it examines the transfer of electrons, which is evidenced in oxidation-reduction (redox) reactions. In our exercise, electrochemistry helps us understand how electric charge leads to the deposition of metals like silver and gold from their ions.
Using a constant electric current, ions in solution are attracted to the charged electrodes, where they gain or lose electrons. This process involves key concepts such as oxidation states, electron transfer, and electrochemical cells.

• Oxidation: Loss of electrons, increasing the oxidation state.
• Reduction: Gain of electrons, decreasing the oxidation state.
• Galvanic Cell: A device that uses spontaneous redox reactions to generate electrical energy.

Electrochemistry enables us to tap into redox reactions like the deposition of metals by monitoring the movement of electrons, crucial for determining oxidation states as in our problem with gold deposition.

Mole Concept

The mole concept is essential in chemistry for quantifying atoms, ions, and molecules. A mole corresponds to Avogadro's number, which is approximately 6.022 x 10²³ particles. In our exercise, we use the mole concept to connect the mass of a substance to the amount of substance in moles.
When calculating the moles of deposited silver and gold, we employ the formula:
\[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]
This calculation converts mass into a tangible number of atoms or ions involved in the reaction, making it indispensable for determining the stoichiometry of reactions.

• Molar Mass: The mass of one mole of a substance in grams, varying for different elements.
• Stoichiometry: The calculation of reactants and products in chemical reactions to ensure a balanced reaction according to the law of conservation of mass.

The mole concept bridges the gap between the atomic scale and the macroscopic world, aiding us in accurately interpreting experimental results like those in the given problem.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are a fundamental aspect of chemical reactions where electron transfer occurs. An oxidation-reduction reaction involves two complementary processes: one substance loses electrons (oxidation) while another gains them (reduction). This dual process is at the heart of balancing chemical reactions.
In our exercise, we observe redox reactions in the formation of silver and gold from their ions:

• Reduction of Silver Ions: \[\text{Ag}^+\,(aq) + e^- \rightarrow \text{Ag}\,(s) \]
• Reduction of Gold Ions:\[\text{Au}^{x+}\,(aq) + x\,e^- \rightarrow \text{Au}\,(s)\]

These equations highlight the electron transfer, where the stoichiometry coefficients indicate the ratio of ions to electrons. Understanding these reactions allows us to ascertain the oxidation state of elements involved, which is critical for identifying the 'x' in gold ions, calculated to be +3 in this exercise.
Such redox equations emphasize the essence of balancing charges and matter, crucial for both electrochemical and traditional chemical reactions.