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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

Predict whether the following reactions would occur spontaneously in aqueous solution at \(25^{\circ} \mathrm{C}\). Assume that the initial concentrations of dissolved species are all \(1.0 M\). (a) \(\mathrm{Ca}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{Cd}(s)\) (b) \(2 \mathrm{Br}^{-}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Br}_{2}(l)+\operatorname{Sn}(s)\) (c) \(2 \mathrm{Ag}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Ni}(s)\) (d) \(\mathrm{Cu}^{+}(a q)+\mathrm{Fe}^{3+}(a q) \longrightarrow$$\mathrm{Cu}^{2+}(a q)+\mathrm{Fe}^{2+}(a q)\)

Calculate the emf of the following concentration cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(0.080 \mathrm{M}) \| \mathrm{Cu}^{2+}(1.2 \mathrm{M})\right| \mathrm{Cu}(s) $$

Which species in each pair is a better oxidizing agent under standard-state conditions? (a) \(\mathrm{Br}_{2}\) or \(\mathrm{Au}^{3+},\) (b) \(\mathrm{H}_{2}\) or \(\mathrm{Ag}^{+}\) (c) \(\mathrm{Cd}^{2+}\) or \(\mathrm{Cr}^{3+},\) (d) \(\mathrm{O}_{2}\) in acidic media or \(\mathrm{O}_{2}\) in basic media.

Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a \(\mathrm{CuSO}_{4}\) solution. During electrolysis, copper at the anode enters the solution as \(\mathrm{Cu}^{2+}\) while \(\mathrm{Cu}^{2+}\) ions are reduced at the cathode. (a) Write half-cell reactions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated with \(\mathrm{Zn}\) and \(\mathrm{Ag} .\) Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain \(1.00 \mathrm{~kg}\) of \(\mathrm{Cu}\) at a current of \(18.9 \mathrm{~A} ?\)

Consider a galvanic cell consisting of a magnesium electrode in contact with \(1.0 \mathrm{MMg}\left(\mathrm{NO}_{3}\right)_{2}\) and a cadmium electrode in contact with \(1.0 \mathrm{M} \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) Calculate \(E^{\circ}\) for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow.
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