Question

A current of \(1.0 \mathrm{~mA}\) is passed through a solution containing \(\mathrm{Ag}^{+}(\mathrm{aq}) .\) Calculate the mass of silver in the solution if all the silver was deposited as Ag metal in \(14.5 \mathrm{~min}\).

Short answer

The mass of silver deposited is approximately 0.000973 g.

Step-by-step solution

Convert Time to Seconds

Convert the time from minutes to seconds because the charge will be calculated in seconds. Given time is 14.5 minutes, multiply by 60 to convert it to seconds: \[ \text{Time in seconds} = 14.5 \times 60 = 870 \text{ seconds} \]

Calculate Total Charge

Use the formula \( Q = I \times t \) where \( Q \) is the total charge, \( I \) is the current in amperes (1 mA = 0.001 A), and \( t \) is the time in seconds. \[ Q = 0.001 \times 870 = 0.87 \text{ C} \]

Calculate Moles of Electrons

Use Faraday's constant, which is approximately \( 96485 \text{ C/mol} \), to calculate the moles of electrons required. \[ \text{Moles of electrons} = \frac{Q}{F} = \frac{0.87}{96485} = 9.02 \times 10^{-6} \text{ mol} \]

Determine Moles of Silver Deposited

For silver deposition, one mole of \( Ag^+ \) requires one mole of electrons. Hence, the moles of silver deposited are equal to the moles of electrons. \[ \text{Moles of Ag deposited} = 9.02 \times 10^{-6} \text{ mol} \]

Calculate Mass of Silver

Use the molar mass of silver, which is approximately \( 107.87 \text{ g/mol} \), to find the mass of silver deposited. \[ \text{Mass of Ag} = \text{Moles of Ag} \times \text{Molar mass of Ag} \] \[ \text{Mass of Ag} = 9.02 \times 10^{-6} \times 107.87 = 0.000973 \text{ g} \]

Key concepts

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis are fundamental in understanding how chemical reactions occur during electrolysis. This process involves passing an electric current through a solution to cause a chemical change. Faraday's first law states that the amount of substance deposited at an electrode is directly proportional to the charge passed through the solution. The second law tells us that the amount of different substances deposited by the same quantity of electricity is proportional to their equivalent weights. These concepts are pivotal in calculations involving the amount of material deposited during electrolysis, helping us bridge the electrical and chemical worlds.

Current and Charge

To grasp the concept of current and charge, it's important to understand their relationship. Current is the flow of electric charge, and its unit is the ampere (A). When a current flows through a solution over time, it moves charged particles like ions, leading to chemical changes. The total charge (Q) can be calculated using the formula \( Q = I \times t \), where \( I \) represents the current and \( t \) the time in seconds. In our example, converting minutes to seconds and multiplying by the current in amperes gives a total charge used in electrolysis. This charge is the key to finding out how much material is deposited.

Molar Mass Calculations

Molar mass calculations involve determining the mass of a substance based on its amount in moles. Molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). In the example, once we know the moles of silver deposited, multiplying by the molar mass of silver gives the total mass of silver deposited. The molar mass essentially acts as a conversion factor between the amount in moles and the mass in grams, making it possible to translate between these two quantities seamlessly. This step is vital in finding the real-world mass of the substance.

Stoichiometry in Redox Reactions

Stoichiometry in redox reactions involves the quantitative relationship between reactants and products during oxidation-reduction reactions. In electrochemistry, stoichiometry helps us understand how much of each substance participates in the reaction. For silver, each Ag^+ ion requires one electron to become a silver atom. This direct one-to-one relationship simplifies calculations, as the moles of electrons equal the moles of silver deposited. By applying stoichiometry, we can predict and verify the amounts of substances used and produced in the reaction, ensuring the calculations align with the chemical reality.