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Investigation of the oxidation states of vanadium

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Introduction

TAS experiment No.17 1. Date 2. Title Investigation of the oxidation states of vanadium 3. Objective Identify the colour of vanadium ion for various oxidation states, which is from +2 to +5. 4. Introduction Vanadium has a range of oxidation states from 0 to +5. The following experiments are designed to identify the colors of vanadium solutions for the various oxidation states, the ease with which the oxidation states can be formed and the stability of the compound once made. 5. Procedures Part A - Various oxidation states of vanadium 1. Prepare a stock solution of 0.1M ammonium metavanadate (NH4VO3) by dissolving the solid in about 25cm3 of 2M sulphuric acid. Record your observation in Table 1. 2. Take out about 3cm3 of the metavanadate solution and slowly add 0.1M iron(II) sulphate solution until there is no further observable change. Record your observation in Table 1. 3. Repeat step 2 with other reagents as outlined in Table 1, and complete the table. Part B - Identification of colours of vanadium compounds 4. According to the standard reduction potentials provided and the results of Part A, deduce the colours of the various oxidation states of vanadium. ...read more.

Middle

It is soluble in water and is colourless solution. So the purple colour of the reaction mixture indicates the presence of vanadium(II) ions and vanadium(II) solution is in more pure form. b) The reaction of ammonium metavanadate and dilute sulphuric acid with tin powder. It's the only reaction of ammonium metavanadate reduces to vanadium(III) ions. Sn2+ is colourless. The reaction mixture only gives out the color of vanadium(III) ions. The physical properties of vanadium (III) ions will not be affected. c) The reaction of ammonium metavanadate and dilute sulphuric acid with aqueous sulphur dioxide. It's because SO42- is colourless while in other reactions which produce VO2+ will give out colored ions. As SO42- is colourless, vanadium(IV) ions can exists in its blue color. (3) Potassium iodide reacts with ammonium metavanadate to produce iodine. The color of iodine is brown/reddish-brown. Then the reaction mixture turns to brown color at first. In order to test the presence of vanadium(IV) ions, it needs to add sodium thiosulphate solution. The sodium thiosulphate solution can reduce iodine to iodide ions which are colourless. The following chemical equation shows the reaction between them. ...read more.

Conclusion

It's accurate to obtain the color change of the mixture. The oxidation state of vanadium is easily oxidized back to vanadium(III) or higher. Air rapidly oxidize the vanadium(II) ions because there are hydrogen ions in the solution. V3+ + e- � V2+ -0.260V 2H+ + 2e- � H2 0V The difference of reduction potentials between two reaction is large. The V2+will be oxidized to V3+and hydrogen ions are reduced to hydrogen. Therefore, some reaction produce V2+ may have color change from purple to green. So the reaction must be covered with cotton wool to prevent air contact with mixtures. The same case occurs when air contact with V3+. Oxygen in air will oxidize the solution to vanadium(IV) state VO2+ ions. Then the solution turns from green to blue. The reaction of ammonium metavanadate and dilute sulphuric acid with tin powder shows blue reaction mixture at last although the reaction produce V3+ions. So the test tube should be covered with cotton wool. 9. Conclusion VO3-, with oxidation state +5, is yellow in color. VO2+, with oxidation state +5, is yellow in color. VO2+, with oxidation state +4, is blue in color. V3+, with oxidation state +3, is green in color. V2+, with oxidation state +2, is purple in color. ...read more.

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