The paper "Two Oxidation States of Vanadium" is a brilliant example of a lab report on chemistry. Vanadium is mostly used to make alloys for tools, construction elements, and engine parts. A mixture of vanadium and gallium can be used to produce superconductive magnets. In industry, vanadium pentoxide is used for ceramics and catalysts. Some vanadium compounds are used to dye fabrics . There is no ore from which vanadium can be recovered. It is a trace element found in rock materials and mining by a product. Some sources are coal, crude oil, and tar sands.
In the United States, for example, vanadium resources are generally associated with uranium ores in sandstone. Often the source of vanadium is the cheaper ferrovanadium from countries such as South Africa, China, Canada, etc. South Africa is also a major supplier of vanadium pentoxide . Vanadium is a rare element belonging to the group of transition metals which means that the valence electrons are in more than one shell. As a result, there are several common oxidation states: +2, +3, +4, and +5 .
The fact that these are readily interconverted led to the invention of the Vanadium Redox Battery which is used for the storage of solar and wind energy . The maximum work obtainable from a reaction is called the Gibbs free energy (∆G ) and is expressed by the following formula: Maximum work = ∆G = -nFE0In this formula ∆G is the Gibbs Free Energy, n is the number of electrons exchanged in the reaction, F is the Faraday constant and E0 is the standard potential. The Gibbs free energy of a reaction determines the nature of this reaction, whether it is spontaneous (∆G< 0) or not.
The purpose of this experiment is to determine the chemistry of the vanadium ions in aqueous solutions containing VO2+ and VO2+. Furthermore, it is of the essence to determine the standard reduction potential for the following chemical reaction: VO2+ (aq) + 2H+ (aq)+ e- → VO2+ (aq) + H2O (l)This result will then enable us to calculate the Gibbs free energy of the reaction in question and conclude whether the reaction is spontaneous. ExperimentFor the reduction of VO2+ we prepared 7 test tubes (13x100mm) and 2 mL of 0.05M VO2+ solution was added to each test tube.
Then the reagents were added as follows: The color of the contents in test tube 1, 4, and 5 changed from yellow to blue and the color of the solution had not changed in test tube 2, 3 and 6. The test tube 2, 3 and 6 were placed in the boiling water bath for 20 minutes. After heating in the boiling water bath, the color of the contents changed from yellow to blue in the test tube 6 and test tube 2 and 3 stayed the same. For the oxidation of VO2+, we prepared 6 test tubes (13x100mm) and 2 ml of 0.05M VO2+ solution was added to each test tube. The color of the contents in the test tube 2 changed from light yellow to dark yellow and the solution of the test tube 1 changed from light yellow to red color.
The color of the solution had not changed in test tube 3, 4, and 5. The test tube 1, 3, 4, and 5 were heated in the boiling water bath for 20 minutes.
With heating, the color of the contents changed from red to yellow in the test tube 1 and light yellow to dark yellow in test tube 4. There were no color changes in test tube 3 and 5. DiscussionIn test tubes 1, 4, 5 and 6 we have recorded a reduction of vanadium, due to the fact that we obtained a colour change from yellow to blue. The reactions are as follows: Test Tube 12 (VO2+ + 2H+ + e- → VO2+ + H2O) E0= xH2SO3 + H2O → SO42- + 2H+ + 2e- E0= -0.20 V2VO2+ + H2SO3 → SO42- +2VO2+ + H2O E0cell= x + (-0.20 V) Test Tube 22 (VO2+ + 2H+ + e- → VO2+ + H2O) E0= x2 H2O → H2O2 + 2H+ + 2e- E0= -1.77 V2VO2+ + 2 H+ → 2VO2+ + H2O2 E0cell = x + (-1.77 V) Test Tube 32 (VO2+ + 2H+ + e- → VO2+ + H2O) E0= x2 Br- → Br2 + 2e- E0= -1.07 V2VO2+ + 2 Br- + 4H+ → 2VO2+ + Br2 + H2O E0cell = x + (-1.07 V)Test Tube 42 (VO2+ + 2H+ + e- → VO2+ + H2O) E0= xHNO2 + H2O → NO32- + 3H+ + 2e- E0= -0.94 VVO2++ HNO2 → NO32- + VO2+ + H+ E0cell = x + (-0.94 V)Test Tube 5VO2+ + 2H+ + e- → VO2+ + H2O E0= xFe2+ → Fe3+ +e- E0= -0.77 VVO2++ Fe2+ + 2H+ → VO2+ + Fe3+ + H2O E0cell = x + (-0.77 V)The reduction in test tube 6 requires heat. 2 (VO2+ + 2H+ + e- → VO2+ + H2O) E0= xH2C2O4 + (heat) → 2CO2- + 2H+ + 2e- E0=-0.492VO2+ + H2C2O4 + (heat) → 2CO2- +2VO2+ + 2H2O E0cell = x + (-0.49 V)In the second part of our experiment, we dealt with the oxidation of vanadium which was successfully accomplished in test tubes 1, 2 and 4 when the yellow solution was obtained.
The chemical reactions can be noted as follows: Test Tube 1H2O2 + 2 H+ + 2e- → 2H2O E0= +1.77 V2(VO2+ + H2O → VO2+ +2H+ + e-) E0= xH2O2 + 2 VO2+→ 2VO2+ 2H+ E0cell = +1.77 V + (-x)Test Tube 2Br2 + 2e- → 2Br- E0= +1.07 V2(VO2+ + H2O → VO2+ +2H+ + e-) E0= xBr2 + 2 VO2+ + H2O → 2VO2+ + 2Br- + 4H+ E0cell = +1.07 V + (-x)Test Tube 3NO3- + 3 H+ + 2e- → HNO2 + H2O E0= +0.94 V2 (VO2+ + H2O → VO2+ + 2H+ + e) E0= x2 VO2+ + H2O + NO3→ 2VO2+ + H+ + HNO2 E0cell =+ 0.94 V + (-x)Test Tube 4Fe3+ + e- → Fe2+ E0= +0.77 VVO2+ + H2O → VO2+ +2H+ + e- E0= xFe3+ + 2 VO2+→ 2VO2+ Fe2+ 2H+ E0cell = +0.77 V + (-x)Test Tube 5SO42- + 4 H+ + 2e- → H2SO3 + H2O E0 = +0.20 V2 (VO2+ + H2O → VO2+ + 2H+ + e) E0= x2 VO2+ + H2O + SO42- → 2VO2+ + H2SO3 E0cell = +0.20V + (-x)ConclusionThe actual strength of oxidizing and reducing agents is characterized by the standard electrode potentials (E0).
These potentials are in turn connected to the Gibbs Free Energy ∆G ( also known by the name of Helmholtz energy), which is a measure of the maximum amount of energy per unit charge available when charge is transferred.
This is expressed in the following formula: Maximum work = ∆G = -nFE0In this formula ∆G is the Gibbs Free Energy, n is the amount of electrons exchanged in the reaction, F is the Faraday constant and E0 is the standard potential. ∆G < 0 (spontaneous reaction) ∆G = 0 (equilibrium) ∆G> 0 (non-spontaneous reaction) This was proven by the experiment.
We have furthermore proven that the oxidation stages of vanadium are readily interconverted. This phenomenon resulted in the invention of the VRB.