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The electrolysis of an aqueous solution of \(\mathrm{NaCl}\) has the overall equation $$ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)+2 \mathrm{OH}^{-}(a g) $$ During the electrolysis, \(0.228 \mathrm{~mol}\) of electrons pass through the cell. (a) How many electrons does this represent? (b) How many coulombs does this represent? (c) Assuming \(100 \%\) yield, what masses of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) are produced?

Short Answer

Expert verified
Based on the given information, calculate the following: (a) The actual number of electrons. (b) The charge carried by these electrons in coulombs. (c) The mass of H2 and Cl2 produced, assuming 100% yield.

Step by step solution

01

Calculate the number of electrons

The number of electrons can be calculated by multiplying the given moles of electrons by Avogadro's number (6.022 x 10^23 particles per mole). In this case: Number of electrons = 0.228 mol × (6.022 x 10^23 electrons/mol) Number of electrons = 1.371 x 10^23 electrons
02

Calculate the charge carried by these electrons

We can find the total charge carried by these electrons by multiplying the number of electrons found in the previous step by the charge of one electron (1.602 x 10^-19 C). Total charge = (1.371 x 10^23 electrons) × (1.602 x 10^-19 C/electron) Total charge = 21,946 C
03

Calculate the mass of H2 and Cl2 produced

Since the yield is 100%, the moles of electrons that passed through the cell (0.228 mol) will be equal to the moles of H2 formed according to the balanced equation: 2 H2O + 2 Cl^- --> H2(g) + Cl2(g) + 2 OH^- Moles of H2 formed = 0.228 mol Moles of Cl2 formed = 0.228 mol Now, we can calculate the mass of each product using their molar masses: Molar mass of H2 = 2.016 g/mol Molar mass of Cl2 = 70.906 g/mol Mass of H2 = moles of H2 × molar mass of H2 Mass of H2 = 0.228 mol × 2.016 g/mol Mass of H2 = 0.460 g Mass of Cl2 = moles of Cl2 × molar mass of Cl2 Mass of Cl2 = 0.228 mol × 70.906 g/mol Mass of Cl2 = 16.207 g So, assuming a 100% yield, 0.460 g of H2 and 16.207 g of Cl2 are produced.

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