arc5005

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How can it be explained that when copper metal is treated with bromine liquid it forms CuBr2, but when it is treated with iodine vapor it forms CuI and not CuI2?


A. I2 has less of a reduction potential than Br2
B. I2 has more of a reduction potential than Br2
C. Iodine salts are more soluble than bromide salts.
D. Iodide prefers to carry a -2 charge while bromide prefers to carry a -1 charge.

A) I2 has less of a reduction potential than Br2
When copper metal is treated with either Br or I, it gets oxidized (loses electrons). With bromine, the copper is oxidzed by two e- to form a dication. WIth iodine, copper is oxidized by 1 e- to form a monocation. Solubility does not explain the difference in the degree of oxidation for copper metal, so C & D are eliminated. Because copper loses more e- when treated with bromine than it does when treated with iodine, it must be that bromine is better at extracting electrons from copper. This means that bromine, Br2, has a greater reduction potential (desire to be reduced) than iodine, I2.

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I'm not sure I understand completely why this is the case. Is it a structural reason? Is it because Iodine forms weaker bonds? Iodine is a stronger acid? Why does Copper lose 2 electrons when treated with bromine, but when treated with iodine it only loses 1 electron?
 

Lawpy

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I'm not sure I understand completely why this is the case. Is it a structural reason? Is it because Iodine forms weaker bonds? Iodine is a stronger acid? Why does Copper lose 2 electrons when treated with bromine, but when treated with iodine it only loses 1 electron?

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:



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: Halogens as Oxidizing Agents

Hope this helps.
 
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