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Chapter 19: Problem 21

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

Expert verified
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

01

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: \(6.00 \mathrm{h}\)
02

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: \[Q = (2.50 \times 10^5 \, \text{A}) (21600\, \text{s}) = 5.4 \times 10^9 \, \text{C}\]
03

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): \[n = \frac{Q}{F}\] \[n = \frac{5.4 \times 10^9\, \text{C}}{96,500\, \text{C/mol}} \approx 55979.3\, \text{mol}\]
04

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: \[n_{\mathrm{Ba}} = \frac{n}{2}\] \[n_{\mathrm{Ba}} = \frac{55979.3\, \text{mol}}{2} = 27989.65\, \text{mol}\]
05

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 \(3,843,757\,\text{g}\).

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