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Chapter 19: Problem 13

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

Short Answer

Expert verified
Yes, \(Fe^{3+}\) can oxidize \(I^{-}\) to \(I_{2}\) under standard-state conditions due to the higher reduction potential of the half reaction \(Fe^{3+} + e^- \rightarrow Fe^{2+}\) compared to the half reaction \(2I^{-} \rightarrow I_{2} + 2e^-\).

Step by step solution

01

Identify the Half Reactions

Recognize that Fe3+ will be reduced to Fe2+ and I- will be oxidized to I2. So we have the following half-reactions: \(Fe^{3+} + e^- \rightarrow Fe^{2+}\) and \(2I^{-} \rightarrow I_{2} + 2e^-\)
02

Determine the Half-Cell Potentials

We can look up the standard electrode potentials for these half-reactions. From tables, we know that: E0(Fe3+/Fe2+) = +0.77 volts and E0(I2/2I-) = +0.54 volts.
03

Check for Spontaneity

Fe3+ has a higher reduction potential than I-. Therefore, Fe3+ will gain electrons, reducing to Fe2+, and I- will lose electrons, oxidizing to I2. The reaction is spontaneous under standard conditions as the reduction potential for the Fe3+/Fe2+ half-cell is greater than the I2/2I- half-cell.

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