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Chapter 19: Problem 46

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

Short Answer

Expert verified
The half-reactions for the electrolysis of molten barium chloride are: \( \mathrm{Ba^{2+}(l)+2e^{-}}\rightarrow \mathrm{Ba(s)} \) and \( \mathrm{2Cl^{-}(l)}\rightarrow \mathrm{Cl_{2}(g)+2e^{-}} \). The amount of barium produced when a current of 0.50 A is supplied for 30 minutes is approximately 0.64 g.

Step by step solution

01

Write half-reactions

Barium chloride (\(\mathrm{BaCl}_{2}\)) in molten state contains \(\mathrm{Ba}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. During electrolysis, reduction and oxidation half-reactions happen. The barium ions are reduced to barium metal and chloride ions are oxidized to chlorine gas. The half-reactions are: \nFor Barium: \( \mathrm{Ba^{2+}(l)+2e^{-}}\rightarrow \mathrm{Ba(s)} \) \nFor Chlorine: \( \mathrm{2Cl^{-}(l)}\rightarrow \mathrm{Cl_{2}(g)+2e^{-}} \)
02

Calculate the quantity of electricity transferred

The quantity of electricity (charge) transferred can be calculated using the formula: \(Q = I \times t\), where \(I\) is current and \(t\) is time. Given that \(I = 0.50 \, A\) and \(t = 30 \, min = 1800 \, s\), we find \(Q = 0.50 \, A \times 1800 \, s = 900 \, C\)
03

Calculate the moles of electrons transferred

The number of moles of electrons (\(n\)) transferred can be found using Faraday鈥檚 law which states that each mole of electrons carries a charge equal to one Faraday. From this, we deduce that \(n = \frac{Q}{F}\), where \(F\) is Faraday鈥檚 constant, approximately \(96485 \, C \, mol^{-1}\). Therefore, \(n = \frac{900 \, C}{96485 \, C \, mol^{-1}} = 0.00933 \, mol\)
04

Calculate the grams of Barium produced

From the half-reaction for barium, it's seen that one mole of barium is produced for every two moles of electrons. Therefore, the moles of barium produced will be \(0.00933 \, mol / 2 = 0.00467 \, mol\). Considering that the molar mass of barium is \(137.3 \, g \, mol^{-1}\), the grams of barium produced can be calculated using the formula: \text{Mass} = \text{Molar mass} \times \text{Moles}. Therefore, the mass produced = \(0.00467 \, mol \times 137.3 \, g \, mol^{-1} = 0.64 \, g\)

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