Question

(a) You desire to plate 5 g of Ag from a solution of \(\mathrm{Ag}^{+}\). Determine the number of amperes required over a period of 1000 seconds. (b) With a current of \(0.5\) ampere, how much time is required to plate out \(10 \mathrm{~g}\) of gold from an \(\mathrm{Au}^{3+}\) solution?

Short answer

For the Ag case, a current of approximately \(447.13 A\) is required over 1000 seconds to plate 5 g of Ag. For the Au case, with a current of 0.5 A, it will take approximately \(2928186.80\) seconds to plate 10 g of Au from an \(\mathrm{Au}^{3+}\) solution.

Step-by-step solution

Write down Faraday's Law of electrolysis

Faraday's Law states that the amount of substance deposited or dissolved at an electrode is directly proportional to the electric charge passed through the solution. Mathematically, the law is represented by the formula: \[m = \frac{Q \times M}{n \times F}\] Where: - \(m\) is the mass of the substance deposited or dissolved (in grams) - \(Q\) is the electric charge passed through the solution (in coulombs) - \(M\) is the molar mass of the substance (in grams/mole) - \(n\) is the number of electrons involved in the electrochemical reaction - \(F\) is the Faraday's constant (approximately \(96485 C/mol\))

Determine the values for the Ag case

In this problem, we are given the following information for part (a): - Desired mass of Ag, \(m = 5 g\) - Time period for plating, \(t = 1000 s\) - Molar mass of Ag, \(M = 107.87 g/mol\) - Silver ions are \(\mathrm{Ag}^+\), which means that only one electron is involved in the electrochemical reaction, so \(n = 1\). We have to find the current (in amperes) required for this case.

Calculate the required charge for Ag

Using the Faraday's law, we can rearrange the formula to solve for the electric charge, \(Q\), needed to deposit the desired mass of Ag. \[Q = \frac{m \times n \times F}{M}\] Plug in the values for Ag: \[Q = \frac{5 \times 1 \times 96485}{107.87}\] \[Q \approx 447133.31 \ \mathrm{C}\]

Calculate the current required for Ag

Since we know the total charge required and the time interval, we can find the current (in amperes) by using the formula: \[I = \frac{Q}{t}\] Plug in the values: \[I = \frac{447133.31}{1000}\] \[I \approx 447.13 \mathrm{A}\] For the Ag case, we require a current of approximately \(447.13 A\) over 1000 seconds to plate 5 g of Ag.

Determine values for the Au case

In part (b), we are given the following information: - Desired mass of Au, \(m = 10 g\) - Current used for plating, \(I = 0.5 A\) - Molar mass of Au, \(M = 197.0 g/mol\) - Gold ions are \(\mathrm{Au}^{3+}\), which means that three electrons are involved in the electrochemical reaction, so \(n = 3\). We have to find the time (in seconds) required to plate the desired mass of Au.

Calculate the required charge for Au

Again, using the Faraday's law, rearrange the formula to solve for the electric charge, \(Q\), needed to deposit the desired mass of Au. \[Q = \frac{m \times n \times F}{M}\] Plug in the values for Au: \[Q = \frac{10 \times 3 \times 96485}{197.0}\] \[Q \approx 1464093.40 \ \mathrm{C}\]

Calculate the time required for Au

Since we know the total charge required and the current, we can find the time (in seconds) by using the formula: \[t = \frac{Q}{I}\] Plug in the values: \[t = \frac{1464093.40}{0.5}\] \[t \approx 2928186.80 \ \mathrm{s}\] For the Au case, with a current of 0.5 A, it will take approximately \(2928186.80\) seconds to plate 10 g of Au from an \(\mathrm{Au}^{3+}\) solution.

Key concepts

Electrochemical Reaction

An electrochemical reaction involves the transfer of electrons between chemical species within a solution.
These reactions are fundamental to electrolysis, which allows for the deposition of metals through electrochemical processes at a cathode.

To understand electrochemical reactions, consider:

• The

  Oxidation and Reduction Processes:

  These occur simultaneously; oxidation involves the loss of electrons, while reduction involves the gain of electrons.
• The

  Role of Electrodes:

  In electrolysis, electrons flow from an anode to a cathode, causing deposited metal. In our exercise, silver ions (Ag⁺) gain electrons to form metallic silver on the electrode.

Understanding this process enables you to apply Faraday's Law of Electrolysis, highlighting the relationship between the number of electrons transferred and the resulting change in oxidation states of the substances involved.

Molar Mass

The concept of molar mass is essential in electrochemical calculations as it connects mass to moles of a substance.
Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).
It is calculated by summing the atomic masses of the elements present in a compound according to its chemical formula.

In electrolysis exercises:

• Importance:

  It is used to determine how much mass corresponds to a given amount of moles, thus linking mass to chemical calculations.
• For example: The molar mass of silver (Ag) is 107.87 g/mol, which is employed in the formula to determine how much silver can be deposited by a certain amount of electric charge.

By understanding molar mass, you can accurately predict how much material will result from electrochemical reactions, making it a vital part of Faraday's Law calculations.

Electric Charge

Electric charge is a fundamental property necessary for electrolysis, powering the movement of electrons through the electrolytic cell.
Electric charge is measured in coulombs (C) and is intrinsic to how much substance can be deposited or dissolved during an electrolysis reaction.

Consider these points:

• Unit of Measurement:

  Electric charge as part of Faraday's Law is calculated using the relationship of current (amperes) and time (seconds): \( Q = I \times t \)

• Role in Calculations:

  For any electroplating process, determining the necessary charge helps in calculating required current or time to plate a specific mass of metal. In our exercise, charge was used to determine the current needed to deposit silver and the time required to deposit gold.

Understanding electric charge allows you to comprehend the electrochemical dynamics and optimize electroplating processes.

Faraday's Constant

Faraday's Constant (

Symbol F

) is a key value in electrochemistry representing the charge of one mole of electrons, approximately 96485 coulombs per mole (C/mol).
It is pivotal in calculating the relationship between electric charge and moles of substance deposited in an electrochemical reaction.

Simplifying with Faraday's Constant, you find:

• Applications:

  It allows calculation of the exact mass of a substance deposited at electrodes during electrolysis when combined with molar mass and number of transferred electrons.

• Reliability and Consistency:

  It standardizes electrochemical calculations, ensuring that whatever the electrolyte or metal, accurate predictions can be made consistently.

In the given exercises, Faraday's constant was integral in applying the formula for Faraday's Law, helping determine the amount of electric charge required for desired mass deposition. Understanding it is crucial within electrolysis, offering reliable and calculable electrochemical reactions.