Question

Consider the electrolysis of molten barium chloride \(\left(\mathrm{BaCl}_{2}\right) .\) (a) Write the half-reactions. (b) How many grams of barium metal can be produced by supplying \(0.50 \mathrm{~A}\) for \(30 \mathrm{~min} ?\)

Short answer

0.641 g of barium metal can be produced.

Step-by-step solution

Identify the Half-Reactions

In the electrolysis of molten barium chloride, \[\mathrm{BaCl}_{2(l)} \rightarrow \mathrm{Ba}_{(s)} + \mathrm{Cl}_{2(g)}\] The half-reaction at the cathode (reduction) is: \[\mathrm{Ba}^{2+} + 2e^- \rightarrow \mathrm{Ba}\] The half-reaction at the anode (oxidation) is: \[2\mathrm{Cl}^- \rightarrow \mathrm{Cl}_{2} + 2e^-\]

Calculate Total Charge Supplied

The total charge (\(Q\)) supplied can be found using the formula: \[Q = I \times t\] where \(I = 0.50 \; \mathrm{A}\) and \(t = 30 \; \mathrm{min} = 1800 \; \mathrm{s}\). Thus, \[Q = 0.50 \; \mathrm{A} \times 1800 \; \mathrm{s} = 900 \; \mathrm{C}\]

Determine Moles of Electrons Transferred

Using Faraday's constant (\(F = 96485 \; \mathrm{C/mol}\)), calculate the moles of electrons:\[\text{moles of } e^- = \frac{Q}{F} = \frac{900 \; \mathrm{C}}{96485 \; \mathrm{C/mol}} \approx 0.00934 \; \mathrm{mol}\]

Calculate Moles of Barium Produced

From the cathode half-reaction: \[\mathrm{Ba}^{2+} + 2e^- \rightarrow \mathrm{Ba} \] It shows 2 moles of electrons are needed to produce 1 mole of barium. So the moles of barium are: \[\frac{0.00934 \; \mathrm{mol} \; e^-}{2} = 0.00467 \; \mathrm{mol} \; \mathrm{Ba}\]

Calculate Mass of Barium Produced

The molar mass of barium is approximately 137.33 \(\mathrm{g/mol}\). Therefore:\[\text{mass} = 0.00467 \; \mathrm{mol} \times 137.33 \; \mathrm{g/mol} \approx 0.641 \; \mathrm{g}\]

Key concepts

Half-Reactions

In electrolysis, the process is broken down into two half-reactions, one occurring at the cathode and the other at the anode. These reactions show the reduction and oxidation processes separately, offering a clearer understanding of the changes taking place.
In the electrolysis of barium chloride (\(\mathrm{BaCl}_{2}\)), the cathode, which is the site of reduction, involves the half-reaction:
\[\mathrm{Ba}^{2+} + 2e^- \rightarrow \mathrm{Ba}\]
This reaction shows barium ions (\(\mathrm{Ba}^{2+}\)) gaining electrons to form barium metal.
Meanwhile, the anode, where oxidation takes place, features the half-reaction:
\[2\mathrm{Cl}^- \rightarrow \mathrm{Cl}_{2} + 2e^-\]
This represents chloride ions (\(\mathrm{Cl}^-\)) losing electrons to form chlorine gas. By breaking down the overall reaction into these specific half-reactions, we gain valuable insight into the electron transfer and chemical species involved in the process.
This understanding is crucial for calculating other aspects such as mass of products formed in electrolysis.

Faraday's Constant

Faraday’s Constant, symbolized as \(F\), plays a crucial role in electrochemical calculations. It represents the charge of one mole of electrons, quantified as approximately \(96485\; \mathrm{C/mol}\) (coulombs per mole). This constant is essential when determining relationships between charge and moles of electrons involved in reactions.
In our barium chloride electrolysis example, understanding it's vital because it allows you to predict how much product forms in response to a given electrical current.

• You start by determining the total charge transferred during the electrolysis, which is derived by multiplying the current (\(I\)) by time (\(t\)).
• This total charge is then divided by Faraday's constant to calculate the moles of electrons transferred.

These moles of electrons serve as a bridge to further calculations, helping deduce how much of each element forms at the electrodes. Thus, Faraday's constant establishes a link between theoretical calculations and practical observations in electrochemistry.

Moles of Electrons

Electrolysis processes heavily rely on the concept, 'moles of electrons', to predict the quantity of substances formed during the reaction. To determine the moles of electrons transferred, you start by calculating the total charge supplied.

• This is achieved through the formula: \(Q = I \times t\), where \(Q\) is the charge, \(I\) is the current, and \(t\) is the time in seconds.
• Once the total charge is known, the moles of electrons can be found using Faraday's constant, \(F = 96485 \; \mathrm{C/mol}\). This calculation is expressed as: \(\text{moles of } e^- = \frac{Q}{F}\)

This conversion helps translate the electrical data into quantifiable chemical data. In the barium chloride example, \(0.00934 \; \mathrm{mol} \; e^-\) were calculated, which is a key step in further calculating the amount of barium produced.
Thus, by understanding moles of electrons, we learn how to connect electric charge with chemical transformations.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is crucial for converting the number of moles of a solid product from an electrochemical reaction into its mass.
For the electrolysis of barium chloride, we focused on barium, which has a molar mass of approximately \(137.33 \; \mathrm{g/mol}\).

• Once you know the moles of barium formed from the reaction, this molar mass enables you to calculate the mass of barium produced by multiplying the number of moles by the molar mass.

In our example, given that \(0.00467 \; \mathrm{mol} \; \mathrm{Ba}\) was formed, multiplying by the molar mass provided the final mass of barium as roughly \(0.641 \; \mathrm{g}\).
This clear-cut calculation shows how molar mass acts as a bridge that transforms mole measurements into real-world, tangible quantities like grams.