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Reduction potential between Iodine & Bromine
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<blockquote data-quote="Lawpy" data-source="post: 19896720" data-attributes="member: 622365"><p>The main reason is because iodide ions are less attracted to water molecules than bromide ions. To understand how strong an oxidizing agent is, it's important to see the steps involved in generating them in aqueous solutions. There are few steps involved in converting halogens into halide ions. Let's use iodine as an example.</p><p></p><p>I2 (s) --> I2 (g) (converting solid iodine to gaseous iodine)</p><p>I2(g) --> 2 I (g) (breaking the iodine-iodine covalent bond to create monatomic iodine atoms)</p><p>2 I (g) + 2 e- --> 2 I- (g) (the iodine atoms gain an electron to become iodide ions --> this is electron affinity)</p><p>2 I- (g) + H2O (l) --> 2 I- (aq) (this is the hydration energy/hydration enthalpy needed to make aqueous iodide ions)</p><p></p><p>So, the iodine reactions sum up to:</p><p></p><p>I2 (s) + 2e- --> 2 I- (aq)</p><p></p><p>This reaction is the half reaction for iodine. And the redox potential for this half reaction depends on the total enthalpy of all the reactions involved in creating it. The total enthalpy here is the sum of the atomization energy (generating gaseous iodine atoms from solid diatomic iodine), electron affinity and hydration energy. This means the lower the absolute value of total enthalpy corresponds to lower oxidizing power. This is related to the following similar equation between free energy and redox potentials:</p><p></p><p><img src="https://image.slidesharecdn.com/04redoxreactionsdissoln-precip-110914014811-phpapp01/95/04-redox-reactionsdissolnprecip-22-728.jpg?cb=1315965921" alt="" class="fr-fic fr-dii fr-draggable " style="" /></p><p></p><p>Experimental data show the main energy component contributing to the differences in total enthalpy and oxidizing power is the hydration enthalpy. Iodide ions are larger than bromide ions and are less attracted to water molecules. The magnitude of hydration enthalpy for iodine is therefore smaller than the magnitude of hydration enthalpy for bromine, so the oxidizing power of iodine is weaker than oxidizing power of bromine.</p><p></p><p>This means iodine is a weaker oxidizing agent than bromine and can only take one electron from copper as opposed to two. This is why CuI forms and CuI2 is unstable.</p><p></p><p>For more information, see here: <a href="https://chem.libretexts.org/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/1Group_17%3A_General_Reactions/Halogens_as_Oxidizing_Agents">Halogens as Oxidizing Agents</a></p><p></p><p>Hope this helps.</p></blockquote><p></p>
[QUOTE="Lawpy, post: 19896720, member: 622365"] The main reason is because iodide ions are less attracted to water molecules than bromide ions. To understand how strong an oxidizing agent is, it's important to see the steps involved in generating them in aqueous solutions. There are few steps involved in converting halogens into halide ions. Let's use iodine as an example. I2 (s) --> I2 (g) (converting solid iodine to gaseous iodine) I2(g) --> 2 I (g) (breaking the iodine-iodine covalent bond to create monatomic iodine atoms) 2 I (g) + 2 e- --> 2 I- (g) (the iodine atoms gain an electron to become iodide ions --> this is electron affinity) 2 I- (g) + H2O (l) --> 2 I- (aq) (this is the hydration energy/hydration enthalpy needed to make aqueous iodide ions) So, the iodine reactions sum up to: I2 (s) + 2e- --> 2 I- (aq) This reaction is the half reaction for iodine. And the redox potential for this half reaction depends on the total enthalpy of all the reactions involved in creating it. The total enthalpy here is the sum of the atomization energy (generating gaseous iodine atoms from solid diatomic iodine), electron affinity and hydration energy. This means the lower the absolute value of total enthalpy corresponds to lower oxidizing power. This is related to the following similar equation between free energy and redox potentials: [IMG]https://image.slidesharecdn.com/04redoxreactionsdissoln-precip-110914014811-phpapp01/95/04-redox-reactionsdissolnprecip-22-728.jpg?cb=1315965921[/IMG] Experimental data show the main energy component contributing to the differences in total enthalpy and oxidizing power is the hydration enthalpy. Iodide ions are larger than bromide ions and are less attracted to water molecules. The magnitude of hydration enthalpy for iodine is therefore smaller than the magnitude of hydration enthalpy for bromine, so the oxidizing power of iodine is weaker than oxidizing power of bromine. This means iodine is a weaker oxidizing agent than bromine and can only take one electron from copper as opposed to two. This is why CuI forms and CuI2 is unstable. For more information, see here: [URL="https://chem.libretexts.org/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/1Group_17%3A_General_Reactions/Halogens_as_Oxidizing_Agents"]Halogens as Oxidizing Agents[/URL] Hope this helps. [/QUOTE]
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