Vanadium Lab Report(teozhiyuan) 1i1x4x

Name: Huong Yieng Kee Class: M04K Teacher’s name: Madam Siti Hani Syed Hamid Subject: Chemistry HL Venue: Chemistry Lab 4, Mara College Banting Date: 8/2/2005 Assessment Criteria: DC, DP

INTERNAL ASSESSMENT CHEMISTRY HIGHER LEVEL Practical 12: The Oxidation State of Vanadium and Manganese INTODUCTION: 1. The electronic structure of vanadium is (Ar) 3d3 4s2. Write the electronic structures of V+, V2+ and V3+. Oxidation States of Vanadium Electronic Structures + V (Ar) 3d3 4s1 V2+ (Ar) 3d3 V3+ (Ar) 3d2 2. Which of these ions have the most stable electronic structure? Explain your answer. V2+ ion has the most stable electronic structure. This is because the the formation of the ions involves the removal of 2 electrons from the outermost 4s subshell. 3. Vanadium(III) compounds are much more common than Vanadium(II) and Vanadium(I) compounds. What factors are responsible for this? The electronic configuration of vanadium(III) is (Ar) 3d2. This means the ion will have one empty 4s orbital and three empty 3d orbitals. The other two 3d orbitals will be singly filled. These altogether 4 empty orbitals will then be bonded to four ligands which will arrange themselves tetrahedrally around vanadium(III) to form a complex ion of charge -2. Since tetrahedral shape is a very stable structure, therefore, vanadium(III) compounds are much more common than vanadium(II) and vanadium(I) compounds. (A)Oxidation States of Vanadium: Data Collection: Reaction 1

Solution Ammonium Vanadate + Sodium hydroxide

2

 + hydrochloric acid

Observation When sodium hydroxide is poured into the beaker that contains ammonium vanadate, the solution of colourless aqueous is formed. When hydrochloric acid is added into the colourless solution, the colour changes from colourless to orange.

3

 + zinc

When zinc is added and the solution is shaken, it changes its colour gradually from orange to blue. After further shaking, the blue colour turned into green. Then, after more shaking, the solution changes its colour to violet (right). All solutions are in their aqueous form.

Table 1: Observation of the process of reduction of Vanadium (V) to Vanadium (II) Num Solution 1 Vanadium (II) + Nitric Acid

Observation The solution changes colour gradually from violet to green to bluish green. Table 2: Observation of the process of oxidation of Vanadium (II)

(B) The oxidation states of Manganese Data collection: (i) The preparation of manganese (VI) compound No. 1 2

3

No. 1 2

3

Solution Observation Potassium manganate(VII), The solution does not change its colour. Its colour stays KMnO4 + sulphuric acid, H2SO4 + purple throughout the experiment. Manganese(IV) oxide, MnO2 Potassium manganate(VII), The solution changes its colour from purple to green in KMnO4 + sodium hydroxide, colour. There is also a brown precipitate formed. NaOH + manganese(IV) oxide, MnO2 The solution changes its colour from green to purple.  + Sulphuric acid Table 3: Observation of the process of preparation of manganese (VI) compound

Solution Hydrated manganese(II) suphate + dilute sulphuric acid + concentrated suphuric acid  + potassium manganate(VII)

Observation The solution does not change its colour. The colour stays colourless.

When potassium manganate(VII) is added, the colourless solution changes its colour to pale pink. Dark brown precipitate is formed inside the pale pink solution. Once the water is added, the solution turns colourless  + water and the dark brown precipitate changes to light brown precipitate. Table 4: Observation of the process of preparation of manganese (III) compound

Data Processing: Part 1 Oxidation states of Vanadium 1. A common vanadium compound is ammonium vanadate, NH4VO3. This isn't very soluble in water and is usually first dissolved in sodium hydroxide solution. 2. The solution can be reduced using zinc and an acid - either hydrochloric acid or sulphuric acid, usually using moderately concentrated acid. 3. The reduction is shown in two stages. Some individual important colours are shown, but the process is one continuous change from start to finish. 4. Vanadium(V) ions can be reduced to vanadium(II) ions by zinc and hydrochloric acid solution. The standard redox potentials for the reactions are as follows: Zn ↔ Zn2+ + 2eEө = -0.76V equation 1 + + 2+ ө VO2 + 2H + e ↔ VO + H2O E = +1.00V equation 2 VO2+ + 2H+ + e- ↔ V3+ + H2O Eө = +0.34V equation 3 3+ 2+ ө V +e ↔V E = -0.26V equation 4 5. The experiment begins with the process of mixing ammonium vanadate(V) with sodium hydroxide and hydrochloric acid. In acid solution, vanadium(V) ions appears as yellow solution. 6. Next is the reduction process of vanadium(V) to vanadium(IV). Zink amalgam is used in this reduction process because of its reducing property. The colour changes from orange to green and lastly to blue. The green colour seen isn't actually another oxidation state. It is just a mixture of of VO2+ and VO2+. From equation 1 and 2, the reaction equation for thhe reduction of vanadium(V) to vanadium(IV) is as number 7. 7. Zn + 2VO2+ + 4H+ ↔ Zn2+ + 2VO2+ + 2H2O 8. After a while, the solution colour changes from blue(VO2+) to green(V3+). The green solution contains vanadium(III) ions. The oxidation number is again reduced from +4 to +3. Equation 1 and equation 3: Zn + 2VO2+ + 4H+ ↔ Zn2+ + 2V3+ +H2O 9. Then, the solution changes its colour from green(V3+) to violet(V2+). The violet solution consists of Vanadium(II) ions. The oxidation number has been reduced from +3 to +2. Equation 1 and equation 4: Zn + 2V3+ ↔ Zn2+ + 2V2+ 10. When the resulting solution is added with nitric acid, the color of the solution will change back from violet to green and lastly to blue. This is because, the nitric acid is an oxidizing agent that will cause the vanadium (II) to be oxidized to vanadium (III) and lastly to vanadium (IV). 11. The colour change and the oxidation states of vanadium in the experiment is summarized as below: Ion Name Oxidation states Colour

VO2+ VO2+ V3+ V2+ Dioxovanadium Oxovanadium Vanadium(III) ion Vanadium(II) ion (V) ion (IV) ion +5 +4 +3 +2 Orange Blue Green Violet Table 5: Colour Change and Oxidation States of Vanadium

Part 2 Oxidation states of Manganese Preparation of manganese (VI) compound 1. MnO4- is a very powerful oxidizing agent. 2. The experiment begins with the mixing of potassium manganate(VII), KMnO4 with dilute sulphuric acid. The mixture produces purple solution, which is the colour of potassium manganate(VIII), KMnO4. In another word, no reaction takes place between these two solutions. 3. The mixture of KMnO4 with sodium hydroxide,however, yields a green solution. This shows the presence of Mn6+ ions in the mixture. 4. Therefore, it could be said that Mn7+ in the mixture had been reduced to Mn6+ by sodium hydroxide. 5. When the green solution which contained Mn6+ ions is added with sulphuric acid, the solution changes its colour from green to purple. 6. Therefore, it could be said that Mn6+ ions had been oxidized to Mn7+ by sulphuric acid. Preparation of manganese (III) compound 1. The mixture of hydrated manganese(II) sulphate with both diluted and concentrated sulphuric acid produced an aqueous solution containing Mn3+ in the form of brown precipitate in pale pink solution. 2. Later, when the pale pink solution of Mn3+ is added with water, it produces a light brown precipitate, indicating that Mn3+ does not dissolve in water readily. Ion Name Oxidatio n states Colour

KMnO4 MnSO4 Potassium Manganese (II) manganate(VII) sulphate +7 +6 Purple

MnO2 Manganese (IV)oxide +4

Mn3+ Manganese(III) ion +3

Green Dark Brown Brown Table 6: The oxidation states of Manganese

Mn2+ Manganese(II) ion +2 Pale Pink