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Chapter 18: Problem 102

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?

Short Answer

Expert verified
To produce \(1.00 \times 10^3 kg\) of Barium through electrolysis of molten barium chloride in 4 hours, first calculate the moles of Barium needed using its molar mass (137.33 g/mol). Then, find the moles of electrons needed for electrolysis, considering that each Barium cation requires 2 electrons. Next, convert the moles of electrons to total charge using the Faraday constant (96485 C/mol). Finally, determine the required current by dividing the total charge by the time (in seconds) for electrolysis. The result will give you the necessary current to achieve this goal.

Step by step solution

01

Determine moles of Barium required

First, we need to find out how many moles of Barium are needed to reach the desired mass of 1000 kg. To do this, we will use the molar mass of Barium (Ba), which is approximately 137.33 g/mol. Given mass of Barium = \(1.00 \times 10^3 kg = 1.00 \times 10^6 g\) Molar mass of Barium (Ba) = 137.33 g/mol Moles of Barium required = \(\frac{Mass}{Molar~mass}\)
02

Calculate needed electrons for electrolysis

In the electrolysis of molten Barium chloride, the reaction for the production of Barium is given as: Ba^2+ + 2e^- -> Ba From the above reaction, we observe that every Barium cation needs 2 electrons for electrolysis. Therefore, we will multiply the moles of Barium by the number of electrons required for electrolysis (2). Moles of electrons needed = moles of Barium × number of electrons per mole of Barium #-}
03

Convert moles of electrons to charge

Now, we'll convert the moles of electrons required into total charge using the Faraday constant (F). The Faraday constant represents the amount of charge per mole of electrons and is equal to 96485 C/mol. Total charge, Q = moles of electrons needed × Faraday constant
04

Determine the time for electrolysis

The given time for electrolysis is 4 hours. We'll convert this time into seconds: Time, t = 4 hours × 3600 seconds/hour
05

Calculate the required current

Now, we can find the required current using the formula: \(Current, I = \frac{Charge}{Time}\) Substitute the values of total charge (Q) and time (t) to obtain the required current. By following these steps, you'll be able to determine the necessary current to produce the desired amount of Barium through electrolysis in the given time.

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Most popular questions from this chapter

The electrolysis of \(\mathrm{BiO}^{+}\) produces pure bismuth. How long would it take to produce \(10.0 \mathrm{~g}\) Bi by the electrolysis of a \(\mathrm{BiO}^{+}\) solution using a current of \(25.0 \mathrm{~A} ?\)

Consider the cell described below: $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M)\right|\left|\mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu} $$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by \(0.20 \mathrm{~mol} / \mathrm{L}\). (Assume \(T=25^{\circ} \mathrm{C} .\) )

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=0.55 \mathrm{~V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

Consider a cell based on the following half-reactions: $$ \begin{aligned} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ} &=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{E}^{\circ} &=0.77 \mathrm{~V} \end{aligned} $$ a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$ \mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q) $$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ The two half-cell reactions are $$ \begin{array}{l} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3} . \mathrm{Ox}-\) ide ions can move through this solid at high temperatures (about \(\left.800^{\circ} \mathrm{C}\right) . \Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is \(-380 \mathrm{~kJ}\). Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.
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