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Chapter 7: Problem 36

The order of ease of oxidation of the ions \(\mathrm{F}^{-}, \mathrm{OH}^{-}\) \(\mathrm{Br}^{-}, \mathrm{SO}_{4}^{2-}\) is (a) \(\mathrm{SO}_{4}^{2-}>\mathrm{F}^{-}>\mathrm{Br}^{-}>\mathrm{OH}^{-}\) (b) \(\mathrm{SO}_{4}^{2-}<\mathrm{Br}^{-}<\mathrm{OH}^{-}<\mathrm{F}\) (c) \(\mathrm{SO}_{4}^{2-}>\mathrm{F}^{-}>\mathrm{Br}^{-}>\mathrm{OH}^{-}\) (d) \(\mathrm{OH}^{-}<\mathrm{Br}^{-}<\mathrm{F}^{-}<\mathrm{SO}_{4}^{2-}\)

Short Answer

Expert verified
To summarize, the correct order of ease of oxidation for the given ions (\(\mathrm{F}^{-}, \mathrm{OH}^{-}, \mathrm{Br}^{-}, \mathrm{SO}_{4}^{2-}\)) is based on their reduction potentials. The order is: \(\mathrm{F}^{-}>\mathrm{Br}^{-}>\mathrm{OH}^{-}>\mathrm{SO}_{4}^{2-}\). This corresponds to option (c).

Step by step solution

01

Recall the definitions of oxidation and reduction potential

Oxidation is defined as the loss of electrons, while reduction is the gain of electrons. The reduction potential of an ion is a measure of how easily it gains electrons. Greater reduction potential means that the ion is more prone to undergo reduction, which also means it's more prone to cause oxidation of other elements or ions.
02

Look up the reduction potentials of the ions

We need to look up the standard reduction potentials of the ions in reduction potential tables. Here are the values for these ions: $$ \mathrm{F}^{-}/\mathrm{F}_{2} : +2.87\ \mathrm{V} \\ \mathrm{OH}^{-}/\mathrm{H}_{2}\mathrm{O} : +0.40\ \mathrm{V} \\ \mathrm{Br}^{-}/\mathrm{Br}_{2} : +1.09\ \mathrm{V} \\ \mathrm{SO}_{4}^{2-}/\mathrm{SO}_{3} : +0.18\ \mathrm{V} $$
03

Determine the order based on reduction potentials

Now that we have the reduction potentials, we can determine the order of ease of oxidation. Remember that higher reduction potential means easier to oxidize. So, we can arrange the ions from highest to lowest reduction potential: 1. \(\mathrm{F}^{-}\): +2.87 V 2. \(\mathrm{Br}^{-}\) : +1.09 V 3. \(\mathrm{OH}^{-}\): +0.40 V 4. \(\mathrm{SO}_{4}^{2-}\): +0.18 V
04

Identify the correct option

Now that we've determined the correct order, we can match our result with one of the options given: \(\mathrm{F}^{-}>\mathrm{Br}^{-}>\mathrm{OH}^{-}>\mathrm{SO}_{4}^{2-}\) This matches option (c): (c) \(\mathrm{F}^{-}>\mathrm{Br}^{-}>\mathrm{OH}^{-}>\mathrm{SO}_{4}^{2-}\) Therefore, the correct answer is option (c).

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