Balancing Redox Reactions. Please answer all parts (Part 8 and Part 9 are what I

Question

Balancing Redox Reactions. Please answer all parts (Part 8 and Part 9 are what I'm most concerned with, the rest is multiple choice.) Thanks!

Potassium permanganate reacts with hydrogen peroxide according to the following UNBALANCED reaction:

MnO4-(aq) + H2O2(aq) O2(g) + Mn2+(aq)

Part 1:

You will use the overall reaction given in the introduction for these problems. First, split the reactions into two half reactions. Which represents the correct separation of these half reactions?

A

MnO4-(aq) O2(g)

H2O2(aq) Mn2+(aq)

B

MnO4-(aq) Mn2+(aq)

H2O2(aq) O2(g)

C

MnO4-(aq) + H2O2(aq)

O2(g) + Mn2+(aq)

Part 2:

We add water molecules to balance the number of oxygens. How many water molecules will you add to the half reaction involving MnO4-? And to which side?

A

4 to the left

B

None

C

2 to the right

D

4 to the right

E

2 to the left

Part 3:

We add hydrogen ions (H+) to balance the number of hydrogens. How many hydrogen ions will you add to the half reaction involving MnO4-? And to which side?

A

None

B

8 to the right

C

4 to the left

D

8 to the left

E

4 to the right

Part 4:

How many water molecules will you add to the half reaction involving H2O2? And to which side?

A

2 to the left

B

None

C

1 to the right

D

2 to the right

E

1 to the left

Part 5:

How many hydrogen ions will you add to the half reaction involving H2O2? And to which side?

A

4 to the right

B

4 to the left

C

2 to the right

D

2 to the left

E

None

Part 6:

We now add electrons to each reaction to balance the charge of on each side of the reaction. How many electrons will you add in the reaction involving MnO4-? And to which side?

A

5 to the left

B

3 to the right

C

None

D

3 to the left

E

5 to the right

Part 7:

We now add electrons to each reaction to balance the charge of on each side of the reaction. How many electrons will you add in the reaction involving H2O2? And to which side?

A

None

B

2 to the left

C

1 to the right

D

1 to the left

E

2 to the right

Part 8:

You will now multiply each ENTIRE half reaction by some coefficient to make both half reactions have the same number of electrons. What will you multiply the half reaction containing MnO4- by? What will you multiply the half reaction containing H2O2 by? Enter a TWO DIGIT number representing how much you multiply MnO4- and H2O2 by, in that order (e.g. if you multiply the MnO4-reaction by 3 and the H2O2 reaction by 2, you would type '32')

Part 9:

Finally, add the two reactions together. Subtract any substance that appears on opposite sides of the reaction and add any substance that appears on the same side of the reaction. Add the coefficients of the final balanced reaction (with lowest whole number coefficients) and enter this number below.

Explanation / Answer

PART 1:

MnO4-(aq) + H2O2(aq) O2(g) + Mn2+(aq)

split in half reactions

MnO4- --> Mn+2

H2O2--> O2

balance O adding H2O

MnO4- --> Mn+2 + 4H2O

H2O2--> O2

balance H adding H+

8H+ + MnO4- --> Mn+2 + 4H2O

H2O2--> O2 + 2H+

balance charges with e-

5e- + 8H+ + MnO4- --> Mn+2 + 4H2O

H2O2--> O2 + 2H+ + 2e-

balance electorons (this is part 9, we simply need 5x2 = 10 electrons, so force that into the equation)

80e- + 16H+ + 2MnO4- --> 2Mn+2 + 8H2O

5H2O2--> 5O2 + 10H+ + 10e-

add both (this is part 9)

5H2O2 + 10e- + 16H+ + 2MnO4- --> 2Mn+2 + 8H2O +5O2 + 10H+ + 10e-

cancel common terms

5H2O2 + 6H+ + 2MnO4- --> 2Mn+2 + 8H2O +5O2

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.