IODINE.


CRYSTALLINE STRUCTURE OF IODINE


Symbol = I

Atomic weight = 127

Molecular weight = 254

Sp.gr. of solid = 4.948

Sp.gr. of vapor = 8.716

Fuses at 113 deg. C.

Boils at 175 deg. C.

Discovered by Courtios in 1811



Occurrence; in combination with Na, K, Ca, and Mg, in sea water, the waters of mineral springs, marine plants and animals; Cod liver oil contains about 37 parts in 100,000.

Preparation; It is obtained from the ashes of seaweed, called kelp or varech. These are extracted with H2O, and the solution evaporated to small bulk. The mother liquid, separated from the other salts, which crystallize out, contains iodides which are decomposed by Cl, aided by heat, and the liberated iodine condensed.

Physical properties; Blue gray, crystalline scales, having a metallic luster. Volatile at all temperatures, the vapor having a violet color and a peculiar odor. It is sparingly soluble in H2O, which, however, dissolves larger quantities on standing over an excess of iodine, by reason of the formation of hydriodic acid. The presence of certain salts, notably potassium iodide, increases the solvent power of H2O for iodine.

Chemical; in its chemical characters I, resembles Cl, Br, but is less active. It decomposes H2O slowly and is a weak bleaching and oxidizing agent. It decomposes hydrogen sulfide with the formation of hydriodic acid and liberation of sulfur. It does not combine directly with oxygen, but dose with ozone. Potassium hydrate solution dissolves it, with formation of potassium iodide and some hypoidite. Nitric acid oxidizes it to iodic acid. With ammonium hydrate solution it forms the explosive nitrogen iodide.

Impurities; Non-volatile substances remain when I is volatilized. Water separates as a distinct layer when I is dissolved in carbon disulfide. Cyanogen iodide appears in white, acicular crystals among the crystals of sublimed I when � Oz. Of the substance is heated over a water bath for 20 min in a cold finger. The last named is the most serious impurity being that it is an active poison.

Analytical characters; Violate vapor. Dissolves in chloroform and carbon disulphide with a violet color. Colors starch paste deep violet blue color, the color disappearing when heated and returning on cooling.


Extracting Iodine from Kelp


  • "Iodine is one of the microelements necessary for human body.
  • "Seawater contains about 0.000005% of iodine which can be absorbed and concentrated by some marine living beings.
  • "There are 1000-4000 mg of iodine existing as iodides in kelp of per thousand grams.
  • "Burning kelp gathers the iodides in ashes.
  • "Decoct the ashes in water, acidify the resultant iodide solution and then evaporate it to dryness.
  • follow-up oxidation with potassium dichromate produces iodine, which is then separated by sublimation.
    1. K2Cr2O7 + 6KI + 7H2SO4 ---> 4K2SO4 + Cr2(SO4)3 + 3I2+ 7H2O
  • Weigh and cut 40 g of commercial dried kelp into small pieces and put them in an iron vessel.
  • Burn the kelp chips with an electric heater in a fume cupboard until all the chips are completely turned to ash.
  • Transfer the ashes to a beaker, add 40 mL of distilled water, heat and boil the suspension, and add a little more water when necessary to make the filtrate be about 30 mL after filtering.
  • Add drop by drop dilute sulfuric acid (2 mol/L) to the filtrate until its value of pH becomes neutral. ( You can use pH paper to check these values. )
  • Put the neutralized filtrate in an evaporating dish, evaporate it by heating to dryness.
  • The residues are parched and ground and uniformly mixed with 2 g of potassium dichromate.
  • Transfer the mixture to a dry tall beaker.
  • Stand a flask containing cold water on the mouth of the beaker.
  • On heating the beaker to sublimate the produced iodine, iodine vapor condenses on the cooler sides of the flask.
  • Stop heating when there is no violet red iodine vapor appearing any more.
  • Collect the obtained iodine crystals.

    LARGE SCALE COLD FINGER
    FOR
    IODINE PERIFICATION


    there are many ways to obtain Iodine crystals. it will depend of the purity you will need and if you want to prepare the Iodine first and then proceed to crystalize it. Iodine is a substance that sublimes easily,that means that once the crystals are formed they will pass even at room temperature to gaseous state quickly and directly without going into a liquid state instead. So in order to keep your crystals you will need an extra special care.

    Ok, lets see an easy experiment supposing you want to prepare the Iodine first:

    You will need:

    Bunsen burner and asbestos square (or a hot plate) Tripod and tin lid Beaker, 250 ml Separating funnel 100ml Evaporating dish Hydrogen peroxide 20 volume Tetrachloromethane or dichloromethane Distilled water 1M solution of Sulphuric acid Supply of ribbon seaweed(Laminaria) - that you can obtain a a biological supply store or from the sea shore Procedure (You must work at a fume cupboard) 1-Collect and dry about a dozen 50cm lenghts of the seaweed and heat them strongly on a tin recipient until they are reduced to ash; probably it will reduce a quite small quantity, about a spoonful. 2- Add to the ash 20 ml of distilled water in a beaker, and heat the suspension until it boils. 3 - Filter the suspension 4 - Acidify the filtered with Sulphuric acid solution 5 - Add then the hydrogen peroxide solution You will observe the formation of a brown color due the iodine liberation from the iodine ions present 6- Transfer the mixture to a separating funnel 7 -Extract the Iodine with tetrachloro methane or other solvent. The result is an organic solution of iodine. 8 - In order to obtain crystals , the solvent may be allowed to evaporate at room temperature, by placing it in an evaporating dish in a fume cupboard. It will result crystals of iodine or rather gray- black brilliant flakes.

    PRODUCTION OF IODINE
    FROM
    SODIUM IODIDE

    For the following experiment you will need to obtain SODIUM IODIDE, IODINE, from tincture of iodine which contains 28 grams liquid, 2% iodine, and 2.5% sodium iodide. To do this you need to evaporate the tincture on a very low heat until nothing is left but sodium iodide, iodine, and H2O.

    At this point you can collect the 2% of iodine by sublimation which would = .56 Gr. from the cold finger, in the bottom of the sublamator you will have 2.5% sodium iodide ( .70 Gr. ). When you have completed this process you sould have 1.26 Gr. iodine if every thing went perfect but in the real world you will have more like 1 Gr. iodine.

    To convert the sodium iodide to iodine from the above sublimation process you can remove it from the sublamtor and apply the following process...

    Chlorine gas is bubbled into a solution of potassium iodide

    Discussion:

    Remember that chlorine "grabs" electrons from other substances. When you mixed chlorine with the potassium iodide solution, the chlorine grabbed the electron it needed from the iodine. The equation for this reaction is:
    Cl2 +2I- ---> 2Cl- + I2

    A test tube contains a layer of potassium iodide solution over a denser layer of carbon tetrachloride. Chlorine is bubbled through the potassium iodide layer. The chlorine reacts with the iodide to form iodine. In the aqueous layer the brown triiodide ion is formed by the reaction of iodine with iodide. In the carbon tetrachloride layer, iodine forms a purple solution. The iodine in the aqueous layer reacts further with chlorine to form iodine monochloride. As more chlorine is added, the iodine reacts with the excess chlorine to form iodine trichloride and the solution de-colorizes. Some iodine remains in the carbon tetrachloride layer.
    Chlorine is bubbled through the potassium iodide layer.


    The chlorine reacts with the iodide to form iodine.


    In the carbon tetrachloride layer, iodine forms a purple solution.


    ( If you are collecting iodine, Stop the chlorine gas at this point, separate the red carbon tetrachloride and wash the H2O solution with more carbon tetrachloride until no more red is visible in the non-polar solvent. You will need to evaporate the carbon tetrachloride with great care due to the fact that iodine will vaporize just above room temperature. If you don't remove the liberated iodine at this point it will be converted to iodine monochloride. Now apply the chlorine gas once more until no more iodine is being liberated into the fresh carbon tetrachloride solution. REMOVING THE IODINE SATURATED CARBON TETRACHLORIDE AS NEEDED).

    Iodine reacts with excess chlorine to form red iodine monochloride.


    Iodine monochloride reacts with excess chlorine to form iodine trichloride...

    ... and the solution decolorizes.



    Discussion:

    This demonstration illustrates that chlorine is a stronger oxidizing agent than iodine, since iodide is oxidized by chlorine to iodine. However, it also illustrates the properties of two inter-halogen compounds, ICl and ICl3. ICl is a ruby-red solid that melts at 27 oC. Its properties are intermediate between those of iodine and chlorine, which makes it similar to bromine. The color at the end of the pictures above is probably due to ICl in the carbon tetrachloride layer, because it is redder than iodine's violet. ICl3 forms a bright yellow solid that consists of I2Cl6 molecules. It decomposes readily to ICl and Cl2, but this is not seen in the video. The color of ICl3 is much lighter than that of ICl or I2, and when it forms the aqueous layer is de-colorized.

    The equations for the reactions that occur during this video are

    2 I-(aq ) + Cl2(aq ) --> I2(aq ) + 2 Cl-((aq ) It's at this point you remove the iodine))

    I-(aq ) + I2(aq ) --> I3-(aq )

    I2(aq ) + Cl2(aq ) --> 2 ICl(aq )

    ICl(aq ) + Cl2(aq ) --> ICl3(aq )



    Electrolysis of potassium iodide to form iodine




    To electrolyze a potassium iodide solution. Electrolysis is a process by which a chemical reaction is carried out by means of the passage)of an electric current. As the reaction proceeds iodide is oxidized at the anode (negative) while water is reduced at the cathode (positive).

    2I--->(aq) I2(s) + 2e-2H2O(l) + 2e- --> 2OH-(aq) + H2(g)

    net reaction: 2I-(aq) + 2H2O(l) -->I2(s) + 2OH-(aq) + H2(g)

    The cathode will be recognized by bubbles (H2) on the electrode and pink phenolphthalein indicator from hydroxide production. Phenolphthalein turns pink in the presence of base and clear in the presence of acid. A slight yellow/brown color to the water around the anode may be observed. This is due to a small amount of iodine being dissolved in the Water.

    Materials:

    alligator clips (4)
    9V battery
    insulated wire
    solid potassium iodide
    spoon
    carbon electrodes ( led removed from two pencils )
    shallow bowl (clear glass or white plastic a beaker would be perfect)
    white paper

    Procedure:

    Add some KI (about the size of your baby fingernail) to the bowl, fill it half full with water and stir until the KI is dissolved. Put a white piece of paper under the bowl if it is a glass bowl.

    Connect the wire to the alligator clips.

    Clip one alligator clip to the tip of the pencils. ( Remove the led from the pencils for use as electrodes ).

    Attach the other ends of the wires to the battery.

    Place the electrodes in the bowl so that the pencil tip is in the solution.

    Observe the electrodes for evidence of reaction. You will observe a pink color to a red color with a saturatted solution.

    To produce iodine for collection you would want to use a saturated solution of sodium iodide.

    How much current will be required to liberate 10 g of iodine from potassium iodide solution in 2 hours ? (Eq. wt. of I2 = 127.0)

    [ Ans. 1.06 amp ]
    Faraday's Laws of Electrolysis

    Faraday's Laws of Electrolysis give the relationship between the amount of material liberated at the electrode and the amount of electric energy that is passed through the electrolyte.

    First Law of Electrolysis : It states that the amount of any substance that is liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.


    W � Q \ Q = I � t

    \ W � I � t

    Therefore, W = Z � I � t


    Where W = Weight of substance deposited or liberated at the electrode

    Z = is the constant (electrochemical equivalent)

    I = current strength in ampere

    t = time in second


    Second Law of Electrolysis : It states that when the same amount of electricity is passed through different electrolytes, the amount of different substances deposited or liberated are directly proportional to the equivalent weight of the substances. Consider two cells connected in a series containing copper sulfate and silver nitrate and if the electric current passes through both the cells then,


    weight of silver deposited � Equivalent weight of silver

    and weight of copper deposited � Equivalent weight of copper

    \?????WAg EAg

    WCu = ECu


    The basic unit of electrical charge is called Faraday which is defined as the charge on one mole of electrons. Electrolysis of sodium iodide solution to find out Faraday can be given as follows :


    Na? (l) + e- � Na (l) Reduction

    2 I - (l) � I2 (s) + 2e- Oxidation

    Net reaction 2Na?(l) + 2I-(l) � 2Na(l) + I2 (s) (Redox)


    The passage of 1 Faraday of charge will produce 1 mole of Sodium metal and 2 Faraday of charge will produce 1 mole of Iodine.


    The passage of 2 Faraday of charge will give 2 moles of sodium metal and 1 mole of iodine.

    ABSTRACT NOTES:
    Preparation:

    Iodine can be prepared in the laboratory by heating potassium Iodide or sodium iodide with dilute sulphuric acid and manganese dioxide.

    2 KI + MnO2 + 3 H2SO4 ==> I2 + 2 KHSO4 + MnSO4 +2 H2O


    Iodine is liberated when a solution of potassium iodide is added to an acidified solution of potassium dichromate.


    Hydrogen Iodide, HI, can be prepared by direct combination of the elements using a platinum catalyst. In the laboratory it is prepared by heating Concentrated Sulfuric Acid, H2SO4, with Sodium Iodide, NaI.



    H2SO4 + 2 NaI ==> 2 HI + Na2SO4



    Potassium iodide from iodized salt

    Objective:

    To remove potassium iodide from iodized salt. Iodine is a halogen. At room temperature it is a bluish-black solid with a metallic luster and is classified as a semiconductor of electricity. Iodine is needed by the thyroid gland during the production of thyroxin, a growth hormone. The thyroid gland obtains the iodine by collecting iodide from the blood plasma and converting it into iodine. A deficiency in iodine will cause the thyroid gland to enlarge (goiter). In order to prevent this table salt has iodine added in the form of potassium iodide (KI) or sodium iodide (NaI). Iodized salt contains 0.01% KI or NaI. The iodine is easily separated from the salt because iodine is soluble in alcohol whereas salt is not.

    Materials:

    filter paper/coffee filter
    steam bath or hot plate
    iodized salt
    3% hydrogen peroxide
    ethanol
    NaI or KI solution
    shallow bowl or plate
    3 small jars with lids (test tubes can be use)
    petroleum ether (alternatives: hexanes, pentane, or diethyl ether)

    Procedure:

    Add 20 g iodized salt and 25 mL ethanol to a jar, tighten lid and shake vigorously. Let the jar sit for 5-10 min and shake occasionally.

    Filter the solution into a shallow bowl or plate and evaporate until dry. A steam bath or hot plate may be needed to quicken the evaporation. (Caution: ethanol is flammable)

    Add 5 mL of 3% hydrogen peroxide to the bowl and warm it slightly until the residue is dissolved.

    Carefully transfer the solution to a small jar, add 1-2 mL of petroleum ether or alternative, tighten the lid and shake. The petroleum ether, hexanes, and pentane should turn slightly (pink due to the presence of iodine. If diethyl ether is used the solution will turn a faint yellow color. Try to use a jar that will allow the diethyl ether to form a thin layer a couple of millimeters thick.

    A standard can be made to compare the color change. Add 5 mL of diethyl ether to 10 mL of hydrogen peroxide in a small jar. Add a few drops of a KI or NaI solution and observe the (colour change in the ether layer.

    NOTE: This is not a viable way to obtain iodine. To liberate 1.12 Grams you would need to process 25 pounds salt. So to reduce 1 Gr. M_ _ _ you would need to process approx. 50 Lbs. Salt.


    MORE NOTES ON THE     Isolation OF IODINE
    Here is a brief summary of the isolation of iodine.
    Iodine is available commercially so it is not normally necessary to make it in the laboratory. Iodine occurs in seawater but in much smaller quantities than chloride or bromide. As for bromine, with suitable sources of brine, it is recovered commercially through the treatment of brine with chlorine gas and flushing through with air. In this treatment, iodide is oxidized to iodine by the chlorine gas.
    2I- + Cl2 2Cl- + I2
    Small amounts of iodine can be made through the reaction of solid sodium iodide, NaI, with concentrated sulphuric acid, H2SO4. The first stage is formation of HI, which is a gas, but under the reaction conditions some of the HI is oxidized by further H2SO4 to form iodine and sulphur dioxide.
    NaI (s) + H2SO4 (l) HI (g) + NaHSO4 (s)
    2HI (g) + H2SO4 (l) I2 (g) + SO2 (g) + 2H2O (l)

    Hydrogen Peroxide Iodine Clock:
    Oxidation of Potassium Iodide by Hydrogen Peroxide
    Description: Two colorless solutions are mixed. After 10 seconds, the colorless mixture suddenly turns blue.
    Concept: Demonstrates a typical clock reaction; shows the effect of the interaction between chemical reactions that have different rates.
    Materials:
    1. Solution A:
    0.6 grams Starch
    30 mLs of Acetic Acid
    4.1 grams of Sodium Acetate
    50 grams of Potassium Iodide
    4.7 grams of Sodium Thiosulfate
    Allow mixture to cool and dilute to 1 liter with distilled water
    1 liter flask
    1. Solution B:
    500 mLs of 3% Hydrogen Peroxide
    500 mLs of distilled water
    1 liter flask
    Safety: Hydrogen Peroxide can be irritating to skin and eyes. Wear safety goggles and gloves.
    Procedure: Mix the two solutions together. Stir by stir bar and stir plate, swirl the mixture in the flask by hand, or mix by transferring the mixture back and forth between the two flasks. Stir until the colorless solution turns blue (about 10 seconds).
    Clean-Up: Remaining blue solution can be washed down drain with water
    .
    Background:
    " The sudden change from colorless to deep blue solutions in this demonstration can be explained with the following sequence of equations[2]:
    3 I-(aq) + H202 + 2 H+ (aq) = I3- (aq) + 2 H20 (l) (1)
    I3- (aq) + 2 S2O3 2- (aq) = 3 I- (aq) + S4O6 2- (aq) (2)
    2 I3- (aq) + starch = starch-I5- complex + I- (aq) (3)
    The first equation indicates that, in an acidic solution, iodide ions are oxidized by hydrogen peroxide to triiodide ions. These triiodide ions are reduced back to iodide ions by thiosulfate ions, as indicated in equation 2. This reaction is much faster than the reaction of equation 1; it consumes triiodide ions as fast as they are formed. This prevents any readily apparent reaction of equation 3. However, after all the thiosulfate ions have been consumed by the reaction of equation 2, triiodide ions react with starch to form the blue starch-pentaiodide complex [3]."


    CRYSTALLINE STRUCTURE OF IODINE

    ORBITAL STRUCTURE OF IODINE





    Hosted by www.Geocities.ws

    1