Procedure : The color of the solution in test tube #2 was recorded. 9mL of the 0

Question

Procedure :

The color of the solution in test tube #2 was recorded. 9mL of the 0.02 potassium permanganate was added to test tubes 3, 4, and 5 and the color of each solution was recorded. 1mL of 2 M sodium hydroxide was added to test tube #3 and the color of the solution was recorded. Then, 1mL of 3 M sulfuric acid was added to test tube #5, and the color of the solution was recorded. Three different pH meters were then attached to test tubes 3, 4, and 5. The pH’s of each solution was recorded. A 50mL beaker was placed on the workbench and then filled with 20 mL of 0.1 M sodium bisulfite. Then a dropper was filled with the 0.1 M sodium bisulfite and added to test tube #3 two drops at a time. After every 2 drops, the reaction was observed and color changes and precipitate formation were recorded. The drops were added until the solution made a subtle color changed and ceased to change color with additional drops. Once this occurred, 2 more drops were added to confirm that the solution has fully reacted. The dropper was discarded into the recycling bin. These steps were repeated for test tubes 4 and 5, starting with filling a dropper with 0.1 M sodium bisulfite. However, drops were added 10 drops at a time for test tube 4 and 20 drops at a time for test tube 5. The station was cleared by adding all test tubes and instruments to the recycling bin

Questions :

1. What is the oxidation half-reaction in test tube #3? Because this reaction is run under basic conditions, you can use HO- and water to balance out oxygen and hydrogen atoms if needed.

a. (HSO_3)^- -> (SO_4)^2- + 3H + 1e^-

b. (HSO_3)^- -> (SO_4)^2- + 2H_2O + 2e^-

c. (HSO_3)^-    -> (SO_4)^2- + H_2O + 1e^-

d. (HSO_3)^-    -> (SO_4)^2- + 3H^+ +4e^-1

2. What was the balanced net-ionic reaction in test tube #3? Because this reaction is run under basic conditions, you can use HO- and water to balance out oxygen and hydrogen atoms if needed. Remember to balance out the electrons in the two half-reactions.

A. HSO3- + 3HO- + 2MnO4- 2MnO42- + SO42- + 2H2O

B. HSO3- + 3HO- + MnO4- MnO42- + SO42- + 2H2O

C. 2MnO4- + H+ + 5HSO3 2Mn2++ 3H2O + 5SO42-

D. 2MnO4- + H+ + 5HSO3 2Mn2++ 3H2O + SO42-

3. What is the oxidation half-reaction in test tube #4?

A. HSO3- + H2O SO42- + 3H+ + 4e-

B. HSO3- + 2HO SO42- + H2O + 2e-

C. HSO3- + H2O SO42- + 3H+ + 2e-

D. HSO3- SO42- + 3H+ + 1e-

4. What was the reduction half-reaction in test tube #4?

A. MnO4- + 1e- MnO42-

B. MnO4- + 8H+ + 5e- Mn2+ + 4H2O

C. MnO4- + 4H+ + 3e- MnO2 + 2H2O

D. MnO4- + 4H+ + 3e- MnO2 + 2H2O + 3e-

5. What was the balanced net-ionic reaction in test tube #4? Because this reaction is run under neutral conditions, you can use water and H+ to balance out oxygen and hydrogen atoms if needed. Remember to balance out the electrons in the two half-reactions.

A. 3HSO3- + 2MnO4-   2MnO2 + 3SO42- + H2O + H+

B. 3HSO3- + MnO4- MnO2 + 3SO42- + H2O + H+

C. 2MnO4- + H+ + HSO3- 2MnO42- + 3H2O + SO42-

D. HSO3- + 2MnO4-   2MnO2 + SO42- + H2O + H+

6.

What is the oxidation half-reaction in test tube #5? Because this reaction is run under acidic conditions, you can use H+ and water to balance out oxygen and hydrogen atoms if needed.

A. HSO3- + H2O SO42-- + 3H+ + 4e-

B. HSO3- SO42- + 3H+ + 1e-

C. HSO3- + H2O SO42- + 3H+ + 2e-

D. HSO3- + 2HO- SO42- + H2O + 1e-

Explanation / Answer

From the given reaction data,

1. Oxidation half-reaction in test tube 3,

B. HSO3- + 2OH- --> SO4^2- + H2O + 2e-

2. Net balanced equation in test tube 3,

A. 2MnO4- + HSO3- + 3OH- --> 2MnO4^2- + SO4^2- + 2H2O

3. Oxidation half-reaction test tube 4,

C. HSO3- + H2O --> SO4^2- + 3H+ + 2e-

4. Reduction half-cell in test tube 4,

C. MnO4- + 4H+ + 3e- --> MnO2 + 2H2O

5. net balanced equation for test tube 4,

A. 3HSO3- + 2MnO4- --> 3SO4^2- + 2MnO2 + H+ + H2O

6. Oxidation half reacion in test tube 5,

C. HSO3- + H2O --> SO4^2- + 3H+ + 2e-

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