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Chapter 18: Problem 139

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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Most popular questions from this chapter

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of \(4.00 \mathrm{~A}\). How long will it take to plate out \(99 \%\) of the platinum in \(0.50 \mathrm{~L}\) of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+}\) ?

A factory wants to produce \(1.00 \times 10^{3} \mathrm{~kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{~h}\) to accomplish this?

The free energy change for a reaction, \(\Delta G\), is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{E}\), for the reaction. This is surprising because \(\mathscr{E}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

Consider the following half-reactions: $$ \begin{aligned} \mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{aligned} $$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 M\) in chloride ion and \(0.020 M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)
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