Sulfur Dioxide And Sulfurous Acid (1959)

[Music] The oxygen compounds of sulfur of significance in general chemistry are those in which the oxidation number of the element is plus4 or plus 6. In this film, we will consider sulfur dioxide and the acid which it forms with water sulfur acid both of which sulfur is present in the oxidation state of 4 plus. Since sulfur has stable veence states both above and below the 4 plus state, sulfur dioxide and its derivatives may serve as either reducing agents or oxidizing agents. Examples of both will be shown, but usually these compounds act as reducing agents. Sulfur dioxide is the principal product formed when sulfur burns as it is now doing in the bottle. And the burning of sulfur constitutes the most direct preparation of the gas. This is the usual method used industrially. In the bottle, a small amount of sulfur is burning in oxygen, and you will note the light blue flame which appears. This bottle is now filled with sulfur dioxide gas and will be set aside for testing later. The most convenient laboratory preparation involves the reaction between sodium hydrogen sulfite which we have placed in this flask and six normal hydrochloric acid. I'll place some of the hydrochloric acid in this dropping funnel. By opening the valve or the stopcock, we can add the acid to the sodium hydrogen sulfide in the flask at any rate that we please. As the acid first enters the flask, the reaction proceeds slowly and the rate may be increased by gentle warming. You can now see bubbles of sulfur dioxide produced. Sulfur dioxide will pass out through the side arm of the flask over into the small Erlin Meyer flask where it's being collected. Sulfur dioxide has a molecular weight of 64 and air as you remember is about 29. So it is entirely feasible to collect this gas by the upward displacement of air. This reaction is more convenient than the preparation of sulfur dioxide by burning sulfur in oxygen because it is under control at all times. With the simple addition of acid or heat, we can cause the reaction to proceed as slowly or as rapidly as we wish. A steady stream of sulfur dioxide is produced. When this flask becomes filled with sulfur dioxide after the air is displaced, we will then use the gas in this flask to test the solubility of sulfur dioxide in water. I will invert this earrer flask with the sulfur dioxide in it. in this beaker of water and agitate to bring the water into contact with the gas. You can see the gas rising inside the flask and the water level going down in the beaker. This experiment indicates that sulfur dioxide is highly soluble in water. Perhaps we shouldn't really say soluble in this case. since the sulfur dioxide is really reacting with the water to form sulfurous acid. But in any case, the gas combines with the water very rapidly as you can see from the flask which is now almost completely filled with water. Just a trace of residual air remains in the flask. Sulfur dioxide is a non-metal oxide and its solutions in water should yield acid. To confirm this suspicion, I will add a few milliliters of water to the bottle in which we burned the sulfur in oxygen. Shaking the bottle will bring more of the gas into contact with the liquid and will speed the solution. I will now put a drop of this solution on a piece of blue litmus paper. The immediate red color which forms indicates that sulfur dioxide does indeed dissolve in water to yield an acid. The equations for the reactions which have been illustrated so far are as follows. First we burn sulfur in oxygen to obtain sulfur dioxide. Next, we prepared sulfur dioxide more conveniently by reacting sodium hydrogen sulfite with hydrochloric acid obtaining sodium chloride, water and sulfur dioxide. Then we saw that sulfur dioxide reacts with water to yield sulfur acid. And finally we saw that sulfur acid ionizes to produce protons and bulfite ions. the proton causing the change in the color of the litmus paper. I will now take uh this flask containing distilled water and place it so that the delivery tube dips beneath the surface of the water in the flask. Sulfur dioxide is so soluble in water that a good deal of it will dissolve in the water by simply causing it to bubble. into the flask in this fashion. We now have a solution of sulfurous acid and using this solution we can illustrate several interesting chemical properties of sulfurous acid. We know that acids react with active metals forming hydrogen and the salt of the metal. to a few milliliters of sulfuric acid solution. In this test tube, I will add a pinch of powdered magnesium. Note the rapid evolution of hydrogen which we can collect and test in the usual way by lighting it. By placing the second test tube above the first, we'll collect some hydrogen and light it. When solutions containing either salt pur acid H2SO3 or the sulfide ion SO3 double minus are oxidized. The product in each case is the sulfate ion. The sulfate ion reacts with the barerium ion. yielding a white precipitate of berium sulfate which is insoluble in hydrochloric acid. The formation of this precipitate serves as a test for the presence of the sulfate ion. In this test tube, we have a few milliliters of our sulfurous acid solution to which we have added a little acetic acid. In the bottle, we have potassium per manganate, a strong oxidizing agent. When uh sulfuric acid, a reducing agent, reacts with oxidizing agents, the product of the reaction is the sulfate ion. As we add the permanganate to the sulfurous acid in the test tube, note the immediate disappearance of the purple color of the permangan. If our prediction is correct, the solution should now contain sulfate ions. To test for sulfate, we will first make the solution strongly acid with hydrochloric acid. And we will then add a few drops of berium chloride. The appearance of a white precipitate of berium sulfate is proof of the presence of the sulfate ion. The precipitate forms immediately and gives every indication that sulfate ion has indeed been formed. The equation for the reaction between potassium per manganate and sulfuric acid may now be derived as follows. We started with potassium per manganate containing the peranganate ion and this was reduced in acid solution to the manganion. Eight hydrogens must be added to the equation to combine with the four oxygen to produce four molecules of water and electrical balance may be achieved by adding five electrons. Sulfur rough acid we saw was oxidized to the sulfate ion. It's necessary to add a molecule of water to obtain the oxygen atom. Four hydrogen atoms or ions and two electrons are produced. Now this equation requires five electrons and this equation produces two. Well, this equation must be multiplied by five and this equation by two so that 10 electrons are involved on each side of the arrow. Adding up the two partial equations, we find that we require two peranganate ions, 16 hydrogen ions, five molecules of sulfur acid and five molecules of water. And we produce two manganos ions, eight molecules of water, five sulfate ions, and 20 hydrogen ions. Hydrogen ions and water molecules appear on both sides of the equation. And we see that we may subtract 16 hydrogen ions from each side, leaving a net of four on the right. And we may subtract five water molecules from both sides, leaving a net of three on the right. The balanced equation for the reaction then reads two per mangan ions and five molecules of sulfuric acid to yield two manganos ions, three molecules of water, five sulfate ions and four hydrogen ions. Another illustration of the reducing action of sulfuric acid is given by its reaction with potassium dromate solution. In the test tube, we have an acidified solution of sulfuric acid. To this solution, we'll add the orange potassium dromate. And you will notice the reduction of the orange dromate ion to the green chromium 3+ or chromic ion. The oxidation product of the sulfuros acid should be the sulfate ion. Again we will acidify the liquid in the tube with hydrochloric acid and then add berium chloride. Our conventional sulfate test appearance of the white precipitate somewhat obscured by the green color of the solution is again a test for the presence of the sulfate ion. We saw that the potassium dromate containing the dromate ion, the orange material was reduced in the reaction to the chromic ion which is green. In order to complete the balancing of this partial equation, we see that we must add 14 hydrogen ions to unite with the seven oxygens to produce seven molecules of water. We must add two chromic ions to account for the two chromiums. And it's necessary to insert six electrons for electrical balance. This is the partial equation for the reduction of the dromate ion to the chromic ion. The oxidation of sulfuric acid involves exactly the same equation as before. We can now write this off without hesitation. In order to balance uh completely, we would now have to multiply this equation by three, combine the two partials, and cancel out hydrogen and water from each side if necessary, exactly as before. The H hallogens may also be used as oxidizing agents for sulfuric acid. To this tube, we will add a few milliliters of bromine water. and then a few milliliters of our solution of sulfuric acid. The red color of the bromine instantly discharged. Now to examine the reaction products from this reaction, we will pour about half of the solution into this tube and leave the remainder in the first tube. The oxidation product of the sulfur acid should be sulfuric acid and this may be again tested for using the hydrochloric acid berium chloride test. As before, the white precipitate indicates that the sulfur acid has been oxidized to the sulfate ion. To this tube, we will add a few milliliters of silver nitrate because the bromine water should have been reduced to the broomemide ion. And if so, we should now form the yellow kurdie precipitate of silver broomemide. I think you can see the The curdy nature of this precipitate has contrasted with the finely divided berium sulfate in the other two. Also, this precipitate is slightly yellow in color. Therefore, sulfuric acid has been oxidized to the sulfate ion as shown by this test. Broine has been reduced to the bromide ion as shown by this test. We saw in the experiment that bromine was reduced to the broomemide ion and sulfuric acid is oxidized to the sulfate ion. It's a simple matter therefore to write the two partial equations and combine them to yield finished equation for the oxidation of sulfuric acid by bromine water. So far all of our reactions have taken place in acid solution. The sulfite ion as contrasted with sulfur acid is a good reducing agent. Also to demonstrate this we will place some of our sulfuros acid in the test tube and add a small quantity of six normal sodium hydroxide until the solution is basic to the litmus test. Hydrogen peroxide is a good oxidizing agent in basic solution and will be used here. To determine whether oxidation of the sulfite ion to the sulfate ion has taken place, we must again make the solution acid with hydrochloric acid. and test with LMAS to make sure that the solution is really acid. And then add berium chloride to make our conventional sulfate ion test. And once again, the immediate formation of the milky white precipitate of berium sulfate indicates that the sulfite ion has been oxidized to the sulfate ion. In basic solution, the sulfur acid reacted with the sodium hydroxide to produce the sulfite ion. Then the sulfide ion was oxidized by the hydrogen peroxide. The half reaction in this case is somewhat different from that from sulfuric acid. But the final product is again the sulfate ion. The peroxide is reduced to two hydroxide ions. And when the partials are combined, this rather simple equation is obtained for the overall reaction between the sulfide ion and hydrogen peroxide. Sulfur dioxide, sulfur acid, and sulfites are all frequently used as bleaches. This cylinder has been filled with sulfur dioxide gas. And we will examine the reaction of this gas with this pink carnation. First, I'll tear off a few petals of the carnation uh and place them out here so that we'll have a record of the original color of the carnation uh before it contacted the sulfur dioxide. Then we'll drop the carnation into the cylinder. And rather rapidly you can see the pink color of the carnation fading to a dead white color. We've chosen a carnation for this experiment because the pigment in pink carnations is one which is unusually susceptible to the bleaching action of sulfur dioxide. But this gas is also used for bleaching the straw used in straw hats. sometimes the paper news used in newspapers and a variety of other purposes. By this time I think you can see that the flower inside the cylinder has turned nearly white as compared with the original pink petals with which we started. So far, all of the reactions studied have shown sulfur in the oxidation state of 4 plus acting as a reducing agent. Sulfur in this oxidation state may act as an oxidizing agent. However, we have filled this bottle with hydrogen sulfide gas and this bottle with sulfur dioxide gas. We'll now place these bottles mouth to mouth. and allow the gases to mingle. As they do so, you can see rather large quantities of elementary sulfur being formed in the reaction system here and being deposited around the insides of the bottle. This sulfur is formed by the oxidation of the hydrogen sulfide by the sulfur dioxide with sulfur being the product of each half reaction. In the reaction we have just seen, hydrogen sulfide and sulfur dioxide both sulfur containing reagents were respectively oxidized and reduced yielding in each case sulfur and water. In this film we have illustrated different methods for the preparation of sulfur dioxide. We have illustrated its solubility in water and the activity of solutions of sulfur dioxide in water namely sulfuric acid as an acid and as a reducing agent. We have also noted the basic oxidation of the sulfite ion. We have seen sulfur dioxide used as the bleach and the reaction of sulfur dioxide as an oxidizing agent with hydrogen sulfide. [Music]