Question

In an electrolysis experiment, a student passes the same quantity of electricity through two electrolytic cells, one containing a silver salt and the other a gold salt. Over a certain period of time, the student finds that \(2.64 \mathrm{~g}\) of Ag and \(1.61 \mathrm{~g}\) of Au are deposited at the cathodes. What is the oxidation state of gold in the gold salt?

Short answer

The oxidation state of gold is +3.

Step-by-step solution

Understanding the Problem

We need to determine the oxidation state of gold (Au) using the masses of Ag and Au deposited, given that equal quantities of electricity pass through both cells.

Calculate the Moles of Ag Deposited

The molar mass of Ag is approximately 107.87 g/mol. Using the formula \( m = nM \), where \( n \) is the number of moles and \( m \) is the mass, calculate the moles of Ag deposited. \[ n(Ag) = \frac{2.64 \, g}{107.87 \, g/mol} \approx 0.0245 \, mol \]

Determine the Charge Passed Through the Silver Cell

For Ag, the ionic reaction is \( Ag^+ + e^- \rightarrow Ag \). Since 1 mole of Ag requires 1 mole of electrons, the charge passed, \( Q \), is \( 0.0245 \, mol \times 1 \, ext{mol}^{-1} \times 96500 \, C/mol \approx 2364.25 \, C \).

Calculate the Moles of Au Deposited

The molar mass of Au is approximately 196.97 g/mol. Calculate the moles of Au deposited. \[ n(Au) = \frac{1.61 \, g}{196.97 \, g/mol} \approx 0.00817 \, mol \]

Determine the Oxidation State of Au from Charge

Assuming \(n\) charges are required to deposit each mole of Au, the relationship is \(n \cdot n(Au) \cdot 96500 = 2364.25 \, C\). Solve for \(n\) using \(\frac{2364.25}{0.00817 \cdot 96500} \approx 3\).

Conclusion

The oxidation state of gold in the gold salt is +3, as 3 electrons are required to reduce each Au ion.

Key concepts

Oxidation State

Oxidation state refers to the number of electrons an atom gains or loses to form a chemical bond. It indicates the charge that an atom would have if all bonds were ionic. In the context of electrolysis, understanding the oxidation state of elements like gold can help predict how many electrons are involved in the reduction or oxidation of the substance.
In our exercise, we determine the oxidation state of gold by calculating how many electrons are needed to deposit gold ions from its salt. This is crucial because the oxidation state directly influences the charge balance in the reaction and the stoichiometry of the resulting chemical equations.

Electrolytic Cells

Electrolytic cells are devices that convert electrical energy into chemical energy through a process called electrolysis. They consist of two electrodes immersed in an electrolyte, where the electric current causes a non-spontaneous chemical reaction. The process involves the movement of ions towards the electrodes, where reduction and oxidation reactions take place.
In the provided exercise, two electrolytic cells are used containing different salts. These cells help demonstrate how the same amount of charge can cause the deposition of different metals by providing capacity to precisely control and measure the amount of electricity passed through the system. This showcases the role of electrolytic cells in facilitating electrochemical reactions.

Molar Mass

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, expressed in grams per mole (g/mol). It allows us to convert between the mass of a substance and the number of moles, providing a link between macroscopic measurements and microscopic entities.
In the exercise, determining the molar mass of silver and gold is essential for calculating the moles of each metal deposited during electrolysis. By dividing the mass of each metal by its molar mass, we gain insight into how much of each element participates in the reaction, enabling further calculations about the electrical charge and chemical changes taking place.

Electrochemical Reactions

Electrochemical reactions include oxidation and reduction processes involving the transfer of electrons between substances. These changes usually occur in electrolytic cells, where electricity facilitates so-called redox reactions.
For example, the reduction reaction happening at the cathode during our experiment involves ion-electron transfer, converting charged ions into neutral metal atoms. By examining these reactions, we can understand how electrical energy converts into chemical energy and vice versa. This deepens our comprehension of how elements change their oxidation state and how energy flow dictates the progression of chemical processes.