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An aqueous solution of \(\mathrm{PdCl}_{2}\) is electrolyzed for \(48.6\) seconds, and during this time \(0.1064 \mathrm{~g}\) of \(\mathrm{Pd}\) is deposited on the cathode. What is the average current used in the electrolysis?

Short Answer

Expert verified
The number of moles of Pd deposited is \(\frac{0.1064 \,\text{g}}{106.42 \,\text{g/mol}}\), and the number of moles of electrons transferred in electrolysis is 2 times the number of moles of Pd deposited. By Faraday's law, the quantity of charge can be found by multiplying the number of moles of electrons with the Faraday constant (\(96485 \,\text{C/mol}\)). To find the average current, divide the total charge by the time (48.6 s). Thus, the average current used in the electrolysis is calculated as follows: Average current (I) = \(\frac{\text{Quantity of charge}}{\text{Time}}\)

Step by step solution

01

Find the number of moles of Pd deposited

First, we need to find the number of moles of Pd deposited on the cathode. We can do this by dividing the mass of Pd deposited (0.1064 g) by its molar mass (106.42 g/mol). Number of moles of Pd = \(\frac{0.1064 \,\text{g}}{106.42 \,\text{g/mol}}\)
02

Calculate the number of moles of electrons transferred in electrolysis

In the electrolysis of PdCl₂, the half-reaction at the cathode is: \(\mathrm{Pd^{2+}} + 2 \mathrm{e^{-}} \rightarrow \mathrm{Pd}\) This means that for each mole of Pd deposited on the cathode, two moles of electrons are transferred. To find the total number of moles of electrons transferred during the electrolysis, we need to multiply the number of moles of Pd deposited by 2. Number of moles of electrons = 2 × (Number of moles of Pd)
03

Use Faraday's law to calculate the average current used in electrolysis

Faraday's law states that the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of charge passed through the solution. This can be expressed as: Quantity of charge (in coulombs) = Moles of electrons × Faraday Constant where the Faraday constant (F) is \(96485 \,\text{C/mol}\). We can use this relation to find the quantity of charge in terms of the number of moles of electrons found in step 2. Once we have the total charge, we can calculate the average current (I) by dividing the total charge by time (t) in seconds. Average current (I) = \(\frac{\text{Quantity of charge}}{\text{Time}}\) Now, plug the values for the moles of electrons, time (48.6 s), and Faraday constant, and calculate the average current in amperes (A).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Molar mass is a fundamental concept in chemistry, which refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Understanding molar mass is essential when converting between the mass of a substance and the number of moles. It allows chemists to relate the weight of a compound to the amount of atoms or molecules it contains. In the context of the exercise, knowing the molar mass of palladium (Pd), which is 106.42 g/mol, enables us to calculate the number of moles of Pd deposited on the cathode. This is done by dividing the mass of Pd deposited by its molar mass:
  • Number of moles of Pd = \( \frac{0.1064 \,\text{g}}{106.42 \,\text{g/mol}} \)
With these calculations, the process of figuring out the amount of a substance in chemical reactions becomes much easier.
Faraday's Law
Faraday's Law of electrolysis is a key principle that links the chemical changes occurring in an electrochemical cell to the flow of electrical charge. It states that the amount of substance processed during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte. The law can be expressed through the formula:
  • Quantity of charge (in coulombs) = Moles of electrons \( \times \) Faraday constant (\( 96485 \,\text{C/mol} \))
In practical applications, such as the exercise above, Faraday's Law helps to calculate how much of a metal like palladium is deposited during an electrolysis process. By understanding this relationship, we can determine the total charge required and, subsequently, the average current used when a known amount of a substance is deposited.
Electrolytic Cell
An electrolytic cell is a device that drives a non-spontaneous chemical reaction through the application of electrical energy. It consists of two electrodes (an anode and a cathode) immersed in an electrolyte - which is a solution or molten compound that conducts electricity. When electricity is passed through the electrolytic cell, ions in the electrolyte move towards the electrodes and undergo chemical reactions, facilitating processes such as deposition or dissolution of materials.
In the exercise, the electrolytic cell setup allows palladium ions (Pd2+) to receive electrons and deposit as Pd metal on the cathode. Understanding the components and functioning of an electrolytic cell is crucial for carrying out various industrial processes including electroplating, electrorefining, and more.
Cathode Reaction
The cathode reaction in an electrolytic process is a reduction reaction where ions from the electrolyte gain electrons. This reaction is crucial because it leads to the deposition of a solid metal from its ion in the electrolyte. In the context of the Pd2+ ions in the exercise, the cathode reaction can be represented as:
  • \( \mathrm{Pd^{2+}} + 2 \mathrm{e^{-}} \rightarrow \mathrm{Pd} \)
This reaction shows that two electrons are needed to reduce Pd2+ ions to metallic palladium. Recognizing cathode reactions helps in understanding the nature of electrochemical deposits and ensures calculations like determining the amount of current or charge involved in the electrolysis are accurate. Studying these reactions allow chemists and engineers to design more efficient and effective electrolytic cells.

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Most popular questions from this chapter

In the electrolysis of a sodium chloride solution, what volume of \(\mathrm{H}_{2}(g)\) is produced in the same time it takes to produce \(257 \mathrm{~L}\) \(\mathrm{Cl}_{2}(g)\), with both volumes measured at \(50 .{ }^{\circ} \mathrm{C}\) and \(2.50 \mathrm{~atm} ?\)

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{M} \mathrm{Ag}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\).a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?\)

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{~V} \end{aligned} $$ In this cell, the silver compartment contains a silver electrode and excess \(\operatorname{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right)\), and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 M\). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\). b. Assuming \(1.0 \mathrm{~L}\) of \(2.0 \mathrm{M} \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons & \\ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned} $$

Hydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. \(\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=0.90 \mathrm{~V}\) \(\mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}\)

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