Copper(II) iodide

[Primarygun]

I heard my teacher say that copper(II) iodide is very unstable and form copper(I) iodide easily, but why?

I've checked the standard electrode potential of

[Math] Cu^{2+}+e^---->Cu^+ [/Math]

[Math] Cu^{2+}+2e^----> Cu [/Math]

I found that the reaction is not spontaneous.

Anyone helps me?

[woelen]

In fact, the redox potentials for I2+2e-->2I(-) and for the copper reactions are fairly close to each other.

What you get is an equlibrium reaction:

2Cu(2+) + 2I(-) (+ 2I(-)) <----> 2Cu(+) + I2 (+ 2I(-)), where the quantities at both sides of the arrow are appreciable. The two I(-) ions between parentheses also are present, because we have 4 iodide ions for two copper (II) ions, so I include them in the equation. These play an important role in the rest of my 'story'.

 

The compound CuI, however, is highly insoluble. So, if some Cu(+) is formed in the equilibrium reaction, then it precipitates as solid CuI with the half of the iodide ions (given in the equation between parentheses) and in that way it is taken out of the equilibrium system. Because the CuI is taken out of the equilibrium system, the reaction is driven to one side, being the side with Cu(+) and I2.

[Primarygun]

then it precipitates as solid CuI with the half of the iodide ions

Does it greatly affect the rate of reaction by the solid's small surface area?

So a formation of solid in an equilibrium plays an vital role, right?

Thanks.

[woelen]

Does it greatly affect the rate of reaction by the solid's small surface area?

So a formation of solid in an equilibrium plays an vital role' date=' right?

Thanks.[/quote']

Indeed, formation of a solid in an equilibrium, based on dissolved species has a VERY large impact on the equilibrium. The same holds for formation of gas, which escapes from solution as bubbles of gas. If such a situation occurs in a chemical reaction, then the reaction will be driven almost completely to the direction, where the gas is present, or where the solid is formed.

 

Another nice example of such a phenomenon is the disproportioning of mercury (I) in the presence of sulfide.

 

The equilibrium involved is Hg2(2+) <----> Hg(s) + Hg(2+). This equilibrium is extremely at the left, even although mercury (0) will form as metal. In fact, before the metal can form macroscopic globules of liquid mercury, it again is made into mercury (I) in the equlibrium.

 

Now, if some H2S is bubbled in, then the situation changes dramatically.

 

H2S also has an equilibrium: H2S <----> H(+) + HS(-) and further:

HS(-) <----> H(+) + S(2-). This eqilibrium also is very much to the left, even the H2S equilibrium is very much to the left.

 

But, when S(2-) and Hg(2+) meet, then such a strong connection is made and such an extremely insoluble product is formed (HgS), that the HS(-) equilibrium is driven to the right en the mercury equlibrium is driven to the right. HgS precipitates from the liquid, Hg metal precipitates from the liquid and the liquid becomes quite acidic, much more than one would expect on the basis of H2S. This is due to the extremely low solubility product of HgS and the great stability of this compound.

[Primarygun]

ndeed, formation of a solid in an equilibrium, based on dissolved species has a VERY large impact on the equilibrium. The same holds for formation of gas, which escapes from solution as bubbles of gas. If such a situation occurs in a chemical reaction, then the reaction will be driven almost completely to the direction, where the gas is present, or where the solid is formed.

We boil hard water to favour the reaction by driving out carbon dioxide, right?

What does hard water exactly mean?

[woelen]

We boil hard water to favour the reaction by driving out carbon dioxide' date=' right?

What does hard water exactly mean?[/quote']

Hard water is water, containing quite some Mg(2+) and Ca(2+) ions, mostly with the bicarbonate ion as counter-ion. In hard water there is the following equilibrium:

 

2HCO3(-) <--> H2O + CO2 + CO3(2-)

 

Calcium bicarbonate and magnesium bincarbonate are reasonably soluble, while calcium carbonate and magnesium carbonate are almost insoluble.

 

By boiling the water, the CO2 indeed is driven out of the liquid and the equlibrium then goes to the right. The concentration of carbonate ion increases and with the Ca(2+) and Mg(2+) ions, a precipitate of CaCO3/MgCO3 is formed. This is what you usually see in cettles, pans and boilers, which frequently are used for boiling hard water.