edited bv GEORGE L. GILBERT

DeniS.on University Granville. OH 43023

Redox Demonshations and Descriptive Chemistry Part 2. Halogens

S U S M ~ BY

Charles E. Ophardt Elrnhurst College Elrnhurst, IL 80128

CHECKED BY

Julle Andrew Cwieton College Northlieid, MN 55057

The descriptive chemistry of hromine and iodine is em- phasized in these two demonstrations by applying the same redox principles as discussed in Part 1.' The names, struc- tu-es, and stability of the oxyanions in acidic or basic solu- tion are descriptive chemistry featured. These demonstra- tions may provide the motivation to look a t the halogen redox reactions. For example, the synthesis of halogens or the various oxyanions may he investigated. In addition, you might make a comparison of the periodic properties and stabilities of various halogen oxidation states.

Dernonstratlon l-Oxldatlon States of Bromine and lodlne

Chemicals and Equipment KBrOs--dissolve 1 gin 88 mL H20 and 12 mL 1 N HzS04 KI-dissolve 1 g in 100 mL HzO Starch-dissolve 1 g starch in 100 mL hoiling water 50 mL distilled water 100-mL graduated cylinder 250-mL beaker

Experimental Procedure The solutions are prepared according to Seger.2 Procedure I-excess 1- ions. In a 250-mL beaker, mix 50 mL

water with 2 mL starch and 1 mL of KI solution. Fill a graduate with 100 mL of KB~OB solution. For the first half of the demonstration, add only a few drops to 1 mL of the KBr03 solution. Note the immediate formation of a blue solution.

Procedure Z-ereess BrOs- ions. In the second half of the demon- stration add all of the remaining KBrOs solution into the heaker. Note a deeper hlue color that gradually turns to purple and finally a pale yellow over a period of 3-4 min.

Reduction potentials for bromine and iodine are

Br03- + 5H+ + 4ec- HBrO + 2H20 Eo = +1.49 V (7)

HBrO + H+ + e- - %Brz + Hz0 EO = +1.59 V (9)

Discussion The redox principle that is applied in all cases in that the

reducing agent must be above the oxidizing agent for a reac-

tion to occur spontaneously. The difference in final products for the two procedures depends upon whether the concen- tration of I- or BrOs- ions is greater. The redox principle that is followed is that all possible reactions will occur until the redox principle is violated.

Procedure I. The choice of I- ion as a reducing agent is easy since I- is the lowest oxidation state possible. I t is also not difficult to establish hromate, BrO3-, ion as the oxidizing agent since i t is in the highest oxidation state listed in the series. As a first guess, hromate ions in half reaction 8 react with I- ions in half reaction 1.

2Br03- + 101- + 12H+ - Brz + 51% + 6Hz0 (11)

Bromate ions could also have reacted with I- in half reac- tion 3 to produce Brz and 103- ions. However, since I- ions are in excess, further reactions between I- and 103- and other products will occur until there are no oxidizing agents left to react, such as,

(1) and (3) - 61- + lo3- + 6H+ -1- + 319 + 3Hz0

(1) and (4) - 41- + 10s- + 6H+ - HI0 + 212 + 3H20

(1) and (5) - 101- + 210~-+ 12H+- 61% + 6Hz0

(1) and (6) - 21- + 2HI0 + 2H+ - 212 + 2H20

The above sequence of reactions shows that elemental iodine, Iz, is the final product from KI. Descriptive chemis- try texts indicate that iodine in the presence of I- forms a triiodide complex, Is-, which interacts with starch to form a hlue complex.

However, the discussion is still incomplete without find- ing the final hromine product. Initially, half reaction 8 was used and gave Brz as a product. Iodide ions could have also reacted with hromate ions in half reaction 7 to produce hypohromous acid (HBrO).

41- + Br03- + 5Ht - HBrO + 21% + 2Hz0

The potential series indicates that HBrO is unstable to- ward disproportionation and will react with itself as both a reducing agent and oxidizing reagent in half reactions 7 and 9.

HBrO + 4HBrO - 2Brz + Br03- + 2H20 + Ht

Since iodide ion is in excess, all BrOa- is quickly used up in reaction 11 to make Ip and BIZ. However, is i t possible for excess I- to react with BIZ? The answer is yes, using half reactions 1 and 2.

In summary, this analysis shows that Iz and Br- ions are the final products when KI is used in excess.

Procedure 2. When potassium bromate is in excess, the final products of the reactionare clearly different from when KI is in excess. The initial reaction (eq 11) is probably the same as previously to produce Iz and BIZ. However, in this case the I- ions are quickly depleted before I- ions can react with Br2 as in the previous final reaction (eq 12). If any

' Ophardt, C. J. Chem. Educ. 1987, 64, xxx. Seger, W. J. Chem. Educ. 1931,8, 166.

Volume 64 Number 9 September 1987 807

HBrO is formed, it disproportionates hack to Brz and BrOs- using half reactions 7 and 9. Both of the ahove reactions leave Brz as the only viable final product.

What happens to the Iz since the blue color with the starch disappears? With Br03- ions as an oxidizing agent in excess, half reactions with 12 as a reducing agent should be found to include

(8) and (5) - Br03- + 12 - Brz + 103-

(8) and (6) - Br08-+ Wz + 2Hz0 + Ht - ll2Br2 + 5HIO

(8) and (4) - 4Br03- + 5HI0 - 2Brp + 5103- + 2H20 + Ht

In addition, H I 0 is also unstable to disproportionation, half reactions 4 and 6, to produce Iq and I03-.

Bromate ions are not strong enough oxidizing agents to change 103- ions into periodic acid, HSIOs, in half reaction 10. In summary, then the final products in this reaction with KBr03 in excess are Brz and 103- ions. The Brp in solution causes the pale yellow color observed a t the end of the reac- tion.

Demonslration 2-Oxldatlon States of Iodine

Chemicals and Equipment Kl-dissolve 2 gin 200 mL water Starch-dissolve 1 g in 100 mL boiling water 15 mL 18 M NaOH 40 mL 11.5 M HC1 100 mL distilled water 100-mL graduated cylinder 10-mL graduated cylinder 500-mL Erlenmeyer

Experimental Procedure Procedure 1-reaction of KI and air. In a 500-mL Erlenmeyer

mix 100 mL of KI solution first with 2-3 mL of starch and then with 20 mL concentrated HCI. Stir or shake vigorously for 2-3 min or until the solution turns purple or blue.

Procedure 2-reaction of I p and I-. Then make the solution basic by adding 15 mL 18 M NaOH, andobserve thatthesolution instant- ly becomes colorless again.

Procedure 3-reaction of 1- and lo3-. Finally, acidify the solu- tion azain with 20 mL concentrated HC1, and observe the return of the blue color in the solution.

Reduction ootentials for iodine are

HI0 + Ht + e- - %I2 + Hz0 E n = +1.45V (23)

HSIOs + Ht + 2e- - 103- + 3H20 EO = +1.6 V (24)

Hz02 + 2Ht + 2e- - 2Hz0 EO= +1.77V (25)

Reactions 13-16 and 19 are basic solutions.

Discussion This series of reactions can he used to show the instability

of oxidation states toward oxygen and disproportionation reactions in either acidic or basic solution. In addition, the relative order of the half reactions changes from acidic to basic solution.

Procedure I . In Procedures 1-3 no apparent oxidizing or reducing agents appear to he added to the solution-only pH changes.

When only I- ions are present in an acid solution, a reac- tion takes place with oxygen dissolved in the water. Reaction 22 with O2 as an oxidizing agent occurs with I- as the reduc- ing agent in reaction 17.

41- + 0% + 4Ht - 212 + 2Hz0

This renctmn shows that I- i; unstable with respect to reac- tiun a i t h air in an arid solution.'l'he 12 rracrs with the starch to form the blue solution.

Procedure 2. A change to basic conditions causes the im- mediate disappearance of the 12 as evidenced by the now colorless solution. The explanation for this reaction is found by using reduction potentials for various iodine oxidation states in basic solution (as noted above). Several possible half reactions that take into account the presence of IZ and excess I- ions from Procedure 1 are as follows:

eqs 17 and 13: 312 +I- + 60H- - 61- + 108- + 3H20

eqs 17 and 14: Iz + Iz + 40H- - 21-+ 210-+ 2Hz0 (26)

eqs 17 and 16: 212 + 10- + 40H-- 41- + 103- + 2Hz0

There are a few more reactions but the final result for all possible combinations yields a mixture of I- and 103- as indicated ahove. These ions are colorless in solution. Note the disorooortionation reaction of I? in reaction 26 under . . basic conditions.

Procedure 3. Finallv, when the mixture that now contains I- and 103- is again acidified, a reaction between these two ions is now possible if reduction potentials under acidic conditions are again used. The easiest half reaction to see is

eqs 17 and 21: 101- + 2103- + 12H+- 612 + 6Hz0

Several other possible half reactions include

eqs 17 and 20: 41- + lo3- + 5Hf - 212 + HI0 + 2Hs0

eqs 17 and 23: 2 1 + 2HIO + 2Ht - 212 + 2H20

eqs 20 and 23: 4HI0 + HI0 - 212 + 1 0 ~ - + 2H20 + Ht

The nroduction of I? from these reactions is the cause for the reappearance of the-blue color. Note that hypoiodous acid is unstable toward dis~ro~ort ionat ion under acidic conditions.

The author is wiliingto make available a booklet contain- ing the theoretical redox principles and a series of nine demonstrations for $5.00 to cover reproduction and ship- ping.

808 Journal of Chemical Education