For each of the test tubes #3, #4, and #5, record the following data in the tabl

Question

For each of the test tubes #3, #4, and #5, record the following data in the table below.

Data Analysis

Based on your observations, to what oxidation state did Mn change in each test tube? Fill in the table below. (I am not sure how to compelte this with the information I have--sorry!)

Use the half-reaction method to determine the net-ionic redox reaction between the permanganate ion and the bisulfite ion in test tube #5.

2.) Vanadium is a transition metal that can be found in the four consecutive oxidation states +2 – +5. In aqueous solution, it forms different colored water complexes as shown in the figure above. [V(H2O)6]2+ is lilac, [V(H2O)6]3+ is green, [VO(H2O)5]2+ is blue, and [VO(H2O)5]3+ is yellow. Which compound do you expect to act as a reducing agent and which as an oxidant?

Test Tube #3 Test Tube #4 Test Tube #5 initial color of the solution plum purple plum purple plum purple pH (basic, neutral, or acidic) 13.27 7.00 .13 final color of the solution bright green milky-clear-ish magenta color of the precipitate, if any No precipitate... No precipitate...   No precipitate...

Explanation / Answer

Mn shows purple colour in +7 oxidation state, green in +6 oxidation state, milky in +4 oxidation state, magenta in +2 oxidation state.

So, in test tube 3:

initial oxidation state is : +7

Mn oxidation state: +6

Number of electrons gained: 1

In test tube 4:

initial oxidation state is : +7

Mn oxidation state: +4

Number of electrons gained: 3

In test tube 5:

initial oxidation state is : +7

Mn oxidation state: +2

Number of electrons gained: 5

Reaction between MnO4- and HSO3-

Oxidation od HSO3-

HSO3- + H2O ------> SO42- + 3H+ + 2e-

Reduction of MnO4-

MnO4- + 8H+ + 5e- --------> Mn2+ + 4H2O

For overall equation, multiply oxidation reaction with 5 and reduction reaction with 2, then add both the equation,

we get:

2MnO4- +5HSO3- + H+ ------> 5SO42- + 2Mn2+ + 3H2O

2.) Oxidation state of V in [V(H2O)6]2+ is 2+,

in [V(H2O)6]3+ is 3+,

in [VO(H2O)5]2+ is 4+,

and in [VO(H2O)5]3+ is 5+

Reducing agent: [V(H2O)6]2+, [V(H2O)6]3+, because they can reduce other species and themselves get oxidised to higher oxidation states.

Oxidising agent: [VO(H2O)5]2+, [VO(H2O)5]3+, because they can oxidise other species and themselves get reduced to lower oxidation states.

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