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Chapter 18: Problem 13

Predict whether \(\mathrm{Fe}^{3+}\) can oxidize \(\mathrm{I}^{-}\) to \(\mathrm{I}_{2}\) under standard-state conditions.

Short Answer

Expert verified
No, \(\mathrm{Fe}^{3+}\) cannot oxidize \(\mathrm{I}^{-}\) to \(\mathrm{I}_{2}\) under standard-state conditions.

Step by step solution

01

Identify the given species

Here, we have two species, \(\mathrm{Fe}^{3+}\) and \(\mathrm{I}^{-}\). \(\mathrm{Fe}^{3+}\) has lost three electrons, so it is in an oxidized state. The question is whether it can oxidize \(\mathrm{I}^{-}\) (an iodine ion) into \(\mathrm{I}_{2}\) (molecular iodine).
02

Consult the standard reduction potential table

The next step is to consult the standard reduction potential table for these two species. From the table, we can find out that the reduction potential for \(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\) is +0.77V, meaning it is favorable for \(\mathrm{Fe}^{3+}\) to gain electrons and be reduced to \(\mathrm{Fe}^{2+}\). Similarly, the reduction potential for \(\mathrm{I}_{2} + 2e^- \rightarrow \ 2\mathrm{I}^{-}\) is +0.54V, implying it is favorable for \(\mathrm{I}_{2}\) to be reduced to \(\mathrm{I}^{-}\).
03

Apply the rule for oxidation

The rule states that a species can oxidize another if its reduction potential is lower than that of the other species. Here the reduction potential of \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) (+0.77V) is higher than the reduction potential of \(\mathrm{I}_{2}\) to \(\mathrm{I}^{-}\) (+0.54V). Thus, \(\mathrm{Fe}^{3+}\) cannot oxidize \(\mathrm{I}^{-}\) to \(\mathrm{I}_{2}\). This means that \(\mathrm{Fe}^{3+}\) does not have the power to induce the iodine ion to release its extra electron and form molecular iodine.

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Most popular questions from this chapter

For each of the following redox reactions, (i) write the half-reactions, (ii) write a balanced equation for the whole reaction, (iii) determine in which direction the reaction will proceed spontaneously under standard-state conditions: (a) \(\mathrm{H}_{2}(g)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow\) \(\quad \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(g)\) (in acid solution) \(\begin{array}{ll}\text { (c) } \mathrm{Cr}(s)+\mathrm{Zn}^{2+}(a q) & \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Zn}(s)\end{array}\)

For a number of years it was not clear whether mercury(I) ions existed in solution as \(\mathrm{Hg}^{+}\) or as \(\mathrm{Hg}_{2}^{2+}\). To distinguish between these two possibilities, we could set up the following system: $$\mathrm{Hg}(l) \mid \text { soln } \mathrm{A} \| \operatorname{soln} \mathrm{B} \mid \mathrm{Hg}(l)$$ where soln A contained 0.263 g mercury(I) nitrate per liter and soln \(\mathrm{B}\) contained \(2.63 \mathrm{~g}\) mercury \((\mathrm{I}) \mathrm{ni}-\) trate per liter. If the measured emf of such a cell is \(0.0289 \mathrm{~V}\) at \(18^{\circ} \mathrm{C},\) what can you deduce about the nature of the mercury(I) ions?

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