Discussion

Date: 2015-10-07; view: 487.

In the first experiment, the change of colour of the solution to deep black blue, caused by starch, indicates that final solution had iodine, which means that in this reaction iodide was oxidized to iodine. Then the half-equation A (I₂ + 2 e^- 2 I^-) moves from right to left and the balanced ionic equation for the reaction is 2 Fe³⁺ + 2 I^- 2 Fe²⁺ + I_2­. From the obtained data it can be said that F > A, because Fe³⁺ is oxidizing agent.

In the second experiment, as the final solution was basic and the test for Fe³⁺ was positive, then the half-equation F (Fe³⁺ + e^- Fe²⁺) moves from right to left and C half-equation (ClO^- + H₂O + 2 e^- Cl^- + 2 OH^-) moves from left to right. And the balanced ionic equation is ClO^- + 2 Fe²⁺ + H₂O Cl^- + 2 OH^- + 2 Fe³⁺. As Cl⁺ was reduced to Cl^-, it is oxidizing agent and it can be stated that C > F and the position of C relative to A and F is C > F > A.

In the third experiment, kerosene's gain of light yellow colour indicates presence of bromine in the final solution, which means that chlorate (I) ions oxidize bromide to bromine, where chlorate is strong oxidizing agent. So, the half-equation E (Br₂ + 2 e^- 2 Br ^- ) moves from right to left and the balanced ionic equation is ClO^- + 2 Br ^- + H₂O Br₂ + Cl^- + 2 OH. And here, C > E.

In the fourth experiment, from the given ionic equation H₂SO₃ + I₂ 2 I^- + H₂SO₄ it can be predicted that kerosene layer would not change its colour as there is no iodine in the final solution, which means that I₂ was reduced and acted as oxidizing agent, so A > B, where A is I₂ + 2e ^- 2I^- and B is SO₄^2- + 4H⁺ + 2e^- H₂SO₃ + H₂O

In the fifth experiment, it can be predicted that if chlorine water is added to solution of bromide ions, then after adding kerosene to the solution it would change its color as Br₂ would present and as Cl has more electronegativity than Br, the reaction should go. The half-equation E would move from right to left 2Br ^- Br₂ + 2e^- and the reaction would be Cl₂ + 2Br ^- 2Cl^- + Br₂. Here, D > E because Cl₂ is reduced and acts as oxidizing agent.

In the sixth experiment, after adding starch to the solution, light yellow colour changed to dark blue, what means that iodine was formed in the final solution and chlorate is oxidizing agent.

In the seventh experiment, after putting glowing splint into a test tube, flashes were produced, what indicates the production of O₂, so H half-equation describes the behaviour of hydrogen peroxide and it moves right to left - H₂O₂ 2H⁺ + O₂. So, here, C > H because sodium chlorate is oxidizing agent and the balanced ionic equation is ClO^- + H₂O + H₂O₂ Cl^- + 2OH^- + 2H⁺ + O₂.

In the eighth experiment, after adding kerosene to the test tube, kerosene gained red colour, which indicates the presence of iodine in the final solution, then A half-equation moves from right to left. G half-equation describes the behaviour of H₂O₂ here and goes from left to right. So, G > A as H₂O₂ is oxidizing agent (H₂O₂ + 2H⁺ + 2e^- H₂O). The balanced ionic equation for the reaction is H₂O₂ + 2H⁺ + 2I^- 2H₂O + I_2.

In ninth experiment, after adding H₂O₂ to manganase, oxygen and water were produced, where H₂⁺O₂^-, H₂⁺O^2- and O₂⁰. In this reaction, hydrogen peroxide acts both as oxidizer and reducer. The equation is 2H₂O₂ + 2H⁺ 2H₂O + 2 H⁺ + O₂. This is a disproportionation reaction, a reaction where one substance both oxidizes and reduces.