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Chapter 19: Problem 72

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

Short Answer

Expert verified
Chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) because the reduction potential of chlorine gas to form chloride ions is positive. Conversely, in the electrolysis of an aqueous solution of \(\mathrm{NaF}\), water has a higher tendency to get oxidized to form oxygen gas and \(\mathrm{H^+}\) ions instead of fluoride ions getting reduced to fluorine gas.

Step by step solution

01

Understanding Electrolysis

Electrolysis is a process by which a chemical reaction is driven by the input of electrical energy. In the electrolysis of a salt solution, the ion that is more eager to accept electrons (the one with a more positive reduction potential) will be reduced at the cathode, while the ion that is less willing to accept electrons (the one with a less positive reduction potential) will be oxidized at the anode.
02

Applying the Concept to Chlorine

Looking at the standard electrode potentials, it can be seen that the reduction potential for chlorine gas (\(\mathrm{Cl_2}\)) to form chloride ions (\(\mathrm{Cl^-}\)) is positive. This means that in the electrolysis of \(\mathrm{NaCl}\), the chloride ions are likely to be reduced to form chlorine gas at the cathode. Simultaneously at the anode, water will get oxidized to form oxygen and H+ ions.
03

Applying the Concept to Fluorine

In contrast, the reduction potential for fluorine (\(\mathrm{F_2}\)) to form fluoride ions (\(\mathrm{F^-}\)) is more positive than that for water. This means that in electrolysis of \(\mathrm{NaF}\), instead of the formation of fluorine gas at the anode, water itself gets oxidized to oxygen gas and \(\mathrm{H^+}\) ions. Thus, fluorine gas is not formed.

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Most popular questions from this chapter

A piece of magnesium ribbon and a copper wire are partially immersed in a \(0.1 \mathrm{M} \mathrm{HCl}\) solution in a beaker. The metals are joined externally by another piece of metal wire. Bubbles are seen to evolve at both the \(\mathrm{Mg}\) and \(\mathrm{Cu}\) surfaces. (a) Write equations representing the reactions occurring at the metals. (b) What visual evidence would you seek to show that \(\mathrm{Cu}\) is not oxidized to \(\mathrm{Cu}^{2+} ?\) (c) At some stage, \(\mathrm{NaOH}\) solution is added to the beaker to neutralize the HCl acid. Upon further addition of \(\mathrm{NaOH},\) a white precipitate forms. What is it?

Write the equations relating \(\Delta G^{\circ}\) and \(K\) to the standard emf of a cell. Define all the terms.

Describe an experiment that would enable you to determine which is the cathode and which is the anode in a galvanic cell using copper and zinc electrodes.

One of the half-reactions for the electrolysis of water is $$ 2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g) $$ If \(0.845 \mathrm{~L}\) of \(\mathrm{H}_{2}\) is collected at \(25^{\circ} \mathrm{C}\) and \(782 \mathrm{mmHg}\), how many faradays of electricity had to pass through the solution?

A galvanic cell is constructed by immersing a piece of copper wire in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}\) solution and a zinc strip in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{ZnSO}_{4}\) solution. (a) Calculate the emf of the cell at \(25^{\circ} \mathrm{C}\) and predict what would happen if a small amount of concentrated \(\mathrm{NH}_{3}\) solution were added to (i) the \(\mathrm{CuSO}_{4}\) solution and (ii) the \(\mathrm{ZnSO}_{4}\) solution. Assume that the volume in each compartment remains constant at \(25.0 \mathrm{~mL}\). (b) In a separate experiment, \(25.0 \mathrm{~mL}\) of \(3.00 M \mathrm{NH}_{3}\) are added to the \(\mathrm{CuSO}_{4}\) solution. If the emf of the cell is \(0.68 \mathrm{~V},\) calculate the formation constant \(\left(K_{\mathrm{f}}\right)\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\).
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