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

Short Answer

Expert verified
The average current used in the electrolysis of the aqueous solution of \(\mathrm{PdCl}_{2}\) is approximately 3.97 A.

Step by step solution

01

Calculate moles of Pd deposited

First, let's calculate the moles of Pd deposited on the cathode. To do this, we can use the formula: Moles = mass / molar mass The molar mass of Pd is 106.42 g/mol. Therefore, the moles of Pd deposited can be calculated as: Moles of Pd = 0.1064 g / 106.42 g/mol = \(9.9995 \times 10^{-4}\) mol
02

Calculate the charge required for the deposition

Next, we need to calculate the charge required for the deposition of Pd. According to Faraday's law, the charge required (in Coulombs) can be calculated using the following formula: Charge (C) = Moles x Faraday constant (F) x n Here, n is the number of electrons exchanged per atom in the reaction, and the Faraday constant (F) is 96,485 C/mol. For the electrolysis of \(\mathrm{PdCl}_{2}\), n = 2 since two chloride ions are removed from each Pd atom: Charge (C) = \(9.9995 \times 10^{-4}\) mol x 96,485 C/mol x 2 = 193.16 C
03

Calculate the average current

Finally, we can find the average current (I) by dividing the total charge by the time (in seconds) for which the electrolysis took place: Average current (I) = Charge (C) / Time (s) Given the time of electrolysis as 48.6 seconds: Average current (I) = 193.16 C / 48.6 s = 3.97 A Thus, the average current used in the electrolysis is approximately 3.97 A.

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Key Concepts

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

Faraday's Law of Electrolysis
Understanding Faraday's Law of Electrolysis is essential when studying the principles of electrochemistry. It provides a quantitative relationship between the amount of substance that undergoes a redox reaction at an electrode during electrolysis and the electric charge passing through the electrolyte.

At the core of the law are two important aspects: the quantity of electric charge and the chemical change that results from the flow of this charge. The electric charge is measured in coulombs (C), and the chemical change is observed in the form of substance deposited or dissolved at the electrodes.

Faraday's first law states that the amount of chemical change is proportional to the amount of charge passed through the electrolyte. The second law states that for a given quantity of charge, the mass of elements deposited or dissolved at the electrodes is proportional to their chemical equivalent weights.

For our exercise concerning the electrolysis of \(\mathrm{PdCl}_{2}\), Faraday's law allows us to calculate the number of coulombs required to deposit a certain amount of palladium, considering two electrons ( \(n = 2\)) are involved in the reduction of each Pd²⁺ ion. Using the calculated amount of charge, we can then work backward to determine the average current used during the electrolysis process.
Calculating Moles in Chemistry
Calculating moles is a fundamental skill in chemistry, connecting the microscopic world of atoms and molecules to the macroscopic world we can measure. A mole is a unit that represents a fixed number of particles, similar to how a dozen represents twelve items. Specifically, one mole contains approximately \(6.022 \times 10^{23}\) entities, known as Avogadro's number.

To calculate moles, you need the mass of the substance and its molar mass, which is the mass of one mole of the substance usually expressed in grams per mole \(\text{g/mol}\). The formula for calculating moles is:
\[\text{Moles} = \frac{\text{mass}}{\text{molar mass}}\]
For example, in our PdCl2 electrolysis problem, we use the mass of deposited palladium and the molar mass of palladium to find the number of moles. This step is crucial because it provides a bridge between the mass of a substance and the number of particles or entities—a bridge that's necessary for applying Faraday's laws and connecting with the concept of electric charge.
Average Current Calculation
In physics and electrical engineering, current is the rate at which electric charge flows. The average current during an electrolysis process is a significant quantity because it indicates how quickly electrons are transferred and, consequently, how fast the redox reaction occurs at the electrodes.

The average current can be calculated by dividing the total charge by the time over which this charge was transferred. The formula for average current (I) is expressed as:
\[I = \frac{Q}{t}\]
where \(I\) is the current in amperes (A), \(Q\) is the charge in coulombs (C), and \(t\) is the time in seconds (s).

In the context of our example with PdCl2 electrolysis, after we have calculated the total charge needed to deposit the palladium based on the moles and Faraday's Law, we can then easily find the average current by using the time given for the electrolysis. This final step brings together all the previous calculations to provide a practical measure of how strong the electric current was during the electrolysis process.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \\ \mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ}=-1.20 \mathrm{V} \end{array}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) and the vanadium compartment contains a vanadium electrode and \(V^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with \(0.0800 M \space \mathrm{H}_{2} \mathrm{EDTA}^{2-},\) resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \space \mathrm{K=?}$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{mL} \space \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{V}\). The solution was buffered at a pH of \(10.00 .\) a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K,\) for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell }}\) at the halfway point in the titration.

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{V} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text {cell. }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{C}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 \mathrm{M}\) and \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M}\)

A galvanic cell is based on the following half-reactions: $$ \begin{array}{ll} \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{V} \end{array} $$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} \mathrm{M}\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00\) atm, and a weak acid, HA, at an initial concentration of \(1.00 \mathrm{M}\). If the observed cell potential is \(0.333 \mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

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 atm?

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