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Chapter 20: Problem 1

Write a balanced equation to represent the electrolysis of molten sodium chloride. What volume of \(\mathrm{Cl}_{2}\) at STP is formed at the anode when \(1.00 \mathrm{~g}\) of sodium is formed at the cathode?

Short Answer

Expert verified
Answer: The volume of Cl₂ gas formed is approximately 0.488 L at STP.

Step by step solution

01

Write the half-reactions

First, we write the half-reactions for the electrolysis of molten sodium chloride: 1. Cathode (reduction): \(\mathrm{Na^+} + \mathrm{e^-} \rightarrow \mathrm{Na}\) 2. Anode (oxidation): \(\mathrm{2Cl^-} \rightarrow \mathrm{Cl}_{2} + \mathrm{2e^-}\)
02

Balance the overall equation

Now, we balance the overall equation by adding the two half-reactions. Since the number of electrons is the same in both half-reactions, we can simply add them: \(\mathrm{Na^+} + \mathrm{2Cl^-} + \mathrm{e^-} \rightarrow \mathrm{Na} + \mathrm{Cl}_{2} + \mathrm{2e^-}\) Next, cancel out the electrons on both sides: \(\mathrm{Na^+} + \mathrm{2Cl^-} \rightarrow \mathrm{Na} + \mathrm{Cl}_{2}\) The balanced equation for the electrolysis of molten sodium chloride is: \(\mathrm{2Na^+} + \mathrm{2Cl^-} \rightarrow \mathrm{2Na} + \mathrm{Cl}_{2}\)
03

Calculate the amount of sodium in moles

We are given that \(1.00 \mathrm{~g}\) of sodium is formed at the cathode. To calculate the amount of sodium in moles, we will use the molar mass of sodium (\(22.99 \mathrm{~g/mol}\)): \(\mathrm{Moles~of~Na} = \frac{1.00 \mathrm{~g}}{22.99 \mathrm{~g/mol}} = 0.0435 \mathrm{~mol}\)
04

Calculate the amount of chlorine in moles

From the balanced equation, we see that for every \(2 \mathrm{~mol}\) of sodium produced, \(1 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) is produced. So, we can calculate the moles of \(\mathrm{Cl}_{2}\) produced as follows: \(\mathrm{Moles~of~Cl_{2}} = \frac{1}{2} \cdot \mathrm{Moles~of~Na} = \frac{1}{2} \cdot 0.0435 \mathrm{~mol} = 0.0218 \mathrm{~mol}\)
05

Calculate the volume of chlorine gas at STP

At STP (Standard Temperature and Pressure: \(273.15 \mathrm{~K}\) and \(1 \mathrm{~atm}\)), \(1 \mathrm{~mol}\) of an ideal gas occupies \(22.4 \mathrm{~L}\). Therefore, we can calculate the volume of \(\mathrm{Cl}_{2}\) gas formed at STP as follows: \(\mathrm{Volume~of~Cl_{2}} = \mathrm{Moles~of~Cl_{2}} \times 22.4 \mathrm{~L/mol} = 0.0218 \mathrm{~mol} \times 22.4 \mathrm{~L/mol} = 0.488 \mathrm{~L}\) The volume of \(\mathrm{Cl}_{2}\) gas formed at the anode when \(1.00 \mathrm{~g}\) of sodium is formed at the cathode is approximately \(0.488 \mathrm{~L}\) at STP.

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