Question

What mass of barium is produced when molten \(\mathrm{BaCl}_{2}\) is electrolyzed by a current of \(2.50 \times 10^{5}\) A for \(6.00 \mathrm{h} ?\)

Short answer

The mass of barium produced when molten \(\mathrm{BaCl}_{2}\) is electrolyzed by a current of \(2.50 \times 10^{5}\) A for \(6.00 \mathrm{h}\) is approximately \(3,843,757\,\text{g}\).

Step-by-step solution

List down the given data

: We are given the following data: 1. Current flowing through the electrolyte: \(2.50 \times 10^5 \mathrm{A}\) 2. Time for which the electrolysis is carried out: \(6.00 \mathrm{h}\)

Calculate the total charge passed through the electrolyte

: Using the formula for the total charge passed (Q) which is the product of current (I) and time (t): \[Q = It\] We first need to convert the time given in hours to seconds to match the units of current. \[t = 6.00\,\text{h} \times \frac{3600 \, \text{s}}{1\,\text{h}} = 21600\, \text{s}\] Now, we can calculate the total charge passed: \[Q = (2.50 \times 10^5 \, \text{A}) (21600\, \text{s}) = 5.4 \times 10^9 \, \text{C}\]

Determine the number of moles of electrons transferred

: To find the number of moles of electrons transferred during the electrolysis, we can use Faraday's constant (F), which is approximately \(96,500\, \text{C/mol}\). The number of moles of electrons (n) is the ratio of the total charge passed (Q) to Faraday's constant (F): \[n = \frac{Q}{F}\] \[n = \frac{5.4 \times 10^9\, \text{C}}{96,500\, \text{C/mol}} \approx 55979.3\, \text{mol}\]

Find the number of moles of barium produced

: In the electrolysis of \(\mathrm{BaCl}_{2}\), the balanced half-reaction occurring at the cathode is: \[\mathrm{Ba}^{2+}(l) + 2e^- \rightarrow \mathrm{Ba}(s)\] From the stoichiometry of this half-reaction, we can see that one mole of barium ions accepts two moles of electrons to form one mole of barium: \[1\,\text{mol Ba}^{2+} \rightarrow 2\,\text{mol e}^{-}\] Given the number of moles of electrons transferred, we can find the number of moles of barium produced: \[n_{\mathrm{Ba}} = \frac{n}{2}\] \[n_{\mathrm{Ba}} = \frac{55979.3\, \text{mol}}{2} = 27989.65\, \text{mol}\]

Calculate the mass of produced barium

: Finally, we can calculate the mass of barium produced using the molar mass of barium (M) which is approximately \(137.33\,\text{g/mol}\): \[m = n_{\mathrm{Ba}} \times M\] \[m = (27989.65\,\text{mol})(137.33\,\text{g/mol}) \approx 3,843,757\,\text{g}\] The mass of barium produced when molten \(\mathrm{BaCl}_{2}\) is electrolyzed by a current of \(2.50 \times 10^{5}\) A for \(6.00 \mathrm{h}\) is approximately \(3,843,757\,\text{g}\).

Key concepts

Faraday's Constant

Faraday's constant is a crucial concept in electrochemistry. It represents the electric charge per mole of electrons. This constant is approximately 96,500 coulombs per mole (C/mol). In the context of electrolysis, Faraday's constant is used to calculate the number of moles of electrons transferred during a reaction.

When performing electrolysis, the total charge (in coulombs) passing through the system is divided by Faraday's constant to determine the number of mole electrons moved. For example:

• If you pass 5.4 × 10⁹ C through an electrolyte, you can find out how many moles of electrons were involved by dividing by Faraday's constant:

\[n = \frac{5.4 \times 10^9 \text{ C}}{96,500 \text{ C/mol}} \approx 55,979.3 \text{ mol}\]The use of Faraday's constant thus provides a bridge between quantities of electricity and chemical amounts, as it relates the two directly through the phenomenon of electrolysis.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the calculation of reactants and products in chemical reactions. It ensures that the reaction is correctly balanced and the proportions of reactants to products are conserved.

In the electrolysis of barium chloride (\(\mathrm{BaCl}_2\)), stoichiometry is used to relate the moles of electrons transferred to the moles of barium produced. The half-reaction at the cathode is:

• \(\mathrm{Ba}^{2+}(l) + 2e^- \rightarrow \mathrm{Ba}(s)\)

This indicates that one mole of barium ions requires two moles of electrons for the reduction to barium metal.

Therefore, the total number of moles of barium produced can be calculated from the number of moles of electrons:

• If 55,979.3 moles of electrons are involved, the moles of barium produced would be:

\[n_{\mathrm{Ba}} = \frac{n}{2} = \frac{55,979.3 \text{ mol}}{2} \approx 27,989.65 \text{ mol}\]By using stoichiometry, we ensure the correct ratio of reactants and products. This is essential for accurate calculations in chemical processes like electrolysis.

Barium Production

Barium production through electrolysis is a typical example of a practical application of electrochemistry. During this process, molten barium chloride (\(\mathrm{BaCl}_2\)) undergoes decomposition.

The electrolysis setup involves a flow of electric current through a molten compound, causing chemical changes at electrodes. Barium ions (\(\mathrm{Ba}^{2+}\)) are reduced at the cathode, turning into solid barium metal upon accepting electrons.

The efficiency of barium production is determined by several factors:

• Current: A constant and sufficient electric current is vital for the reaction to proceed over time.
• Electrolyte purity: Pure electrolytes ensure the correct reaction pathway and minimize side-reactions.
• Electrode material: Suitable electrodes help optimize the yield and reduce resource consumption.

After calculating the moles of barium produced, the mass can be determined using the molar mass of barium (137.33 g/mol):

• This helps in predicting how much barium is formed and allows industries to scale this knowledge appropriately for mass production:

\[m = n_{\mathrm{Ba}} \times 137.33 \text{ g/mol} \approx 3,843,757 \text{ g}\]Thus, understanding and precisely calculating the parameters involved in electrolysis is crucial for the efficient and economic production of metals like barium.