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If everything is in a mixture of alcohol and water, would it be possible for magnesium chloride and iodine to react and produce something?

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Magnesium chloride and Iodine with form an equilibrium. This is a bit of a simplification of the reactions but the dissociated chlorides and free iodine will reversibly react to form iodide and triiodide ions and chlorine.
$$\ce{Cl- + 1/2 I2 <=> 1/2 Cl2 + I-} $$ $$\ce{MgCl2 + I2 <=> MgI2 + Cl2} $$ $$\ce{I- + I2 <=> I3-} $$ $$\ce{MgI2 + 2I <=> Mg(I3)2} $$ The amount of products of this reaction though will not be very appreciable, but yes a very literal sense magnesium chloride and iodine will react.

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    $\begingroup$ $\ce{<=>}$ gives you the correct arrows: $\ce{<=>}$ $\endgroup$ – orthocresol Dec 29 '15 at 2:47
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    $\begingroup$ what do you mean by not be very appreciable? Could you quantify it generally? $\endgroup$ – intuition Dec 29 '15 at 4:55
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    $\begingroup$ That would be pretty negligible. On the same grounds you may claim that any two compounds react with each other, which also would be technically true in some sense. $\endgroup$ – Ivan Neretin Dec 29 '15 at 5:52
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    $\begingroup$ Looking at it macroscopically, the first two equilibria are so far on the reactants’ side, that one would say ‘no reaction is observed’. $\endgroup$ – Jan Dec 29 '15 at 13:06
  • $\begingroup$ (-1): Disagree. See my explanation and source. $\endgroup$ – AJKOER Feb 9 at 18:25
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For the record, elemental iodine does not displace chloride!

Per an educational source, for example, to quote:

These reactions will take place in one direction only. The reaction of a halogen X2 with a different sodium halide (NaY) will occur only if X2 is more reactive than Y2. If X2 is less reactive than Y2, the reaction shown in Equation 1 will not take place.

However, in the real world, does this mean that in the presence of a transition metal impurity (say iron in the form of ferrous as found in tap water) is it likely that aqueous $\ce{MgCl2}$ is unreactive with an iodine presence?

To answer, I will explore some advanced fenton-type chemistry (which could be subject to recycling with light).

Consider the reaction between ferrous and HOI formed in the first step with the reaction of iodine and water:

$\ce{I2 (s) + H2O (l) <=> H+ + I- + HOI}$

$\ce{Fe(2+) + HOI -> Fe(3+) + .IOH-}$

And, at pH over 5 (assuming chemistry paralleling that of the chlorine-based $\ce{.ClOH-}$ radical anion):

$\ce{.IOH- -> .OH + I-}$

$\ce{Cl- + .OH -> .Cl + OH-}$ (Source)

$\ce{.Cl + .Cl -> Cl2}$

But, at pH under 5:

$\ce{.IOH- -> .I + OH-}$

implying likely no chlorine creation.

Note, per photolysis:

$\ce{I- + hv -> .I + e-}$

which could recycle iron ions:

$\ce{Fe(3+) + e- -> Fe(2+)}$.

Also, any ferric iodide formed is unstable and reverts to ferrous, which creates a cyclical clock reaction system (discussed here).

Bottom line, in my assessment, except with favorable impurities, pH and light, it is not likely that any free chlorine, as an intermediate, would be formed in a system of MgCl2 and I2.

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