How do i solve the following redox reaction question?

ANSWER:

H2O + 2 KMnO4 + 3 Na2SO3 --> 2 MnO2 + 2 KOH + 3 Na2SO4

*********HOW TO*********

Redox acronym (In terms of electrons)

O.I.L. R.I.G

Oxidation

Is

Loss

Reduction

Is

Gain

MnO2 tells us this is a redox reaction occurring in a basic solution.

You can balance redox reactions in a basic solution in almost the same way as if they were in an acidic solution. Initially we will assume the reaction is taking place in an acidic solution but we'll switch over to basic solution problem solving after balancing hydrogen ions.

**Step 1**

Separate into two 1/2 reactions.

Reduction:

(-)MnO4 --> MnO2

(we drop potassium on the left side because the K-Mn bond will dissociate in solution)

Oxidation:

(2-)SO3 --> (2-)SO4

**Step 2**

Balance all atoms OTHER than O or H (Already done)

**Step 3**

Starting with the reduction reaction, balance O's by adding H2O to the side that is lacking O.

(-)MnO4 --> MnO2 + 2 H2O

**Step 4**

Now balance H+ (NOT H2)

4 H(+) + (-)MnO4 --> MnO2 +2 H2O

**Step 5**

Now we consider the fact that this is in a basic solution by adding the same number of OH- ions as we have H+ ions to both sides of the half reaction.

4 (-)OH + 4 H(+) + (-)MnO4 --> MnO2 + 2 H2O + 4 (-)OH

**Step 6**

Group H+ and OH- to form water (H2O) and cancelling out equal # of waters from both sides.

On the LEFT side: 4 (-)OH + 4 H(+) = 4 H2O

4 H2O + (-)MnO4 --> MnO2 + 2 H2O + 4 (-)OH (Cancel 2 H2O from both sides)

2 H2O + (-)MnO4 --> MnO2 + 4 (-)OH

**Step 7**

Balance charge (electron = e-)

Net charge on the LEFT side = -1

Net charge on the RIGHT side = -4

So the LEFT side needs 3 e-

3 e- + 2 H2O + (-)MnO4 --> MnO2 + 4 (-)OH

(See how e- are GAINED? Gained = reduction)

**Step 8**

Repeat these steps for the oxidation reaction.

(2-)SO3 --> (2-)SO4

H2O + (2-)SO3 --> (2-)SO4

H2O + (2-)SO3 --> (2-)SO4 + 2 H(+)

2 (-)OH + H2O + (2-)SO3 --> (2-)SO4 + 2 H(+) + 2 (-)OH

2 H(+) + 2 (-)OH = 2 H2O

2 (-)OH + (2-)SO3 --> (2-)SO4 + H2O

2 (-)OH + (2-)SO3 --> (2-)SO4 + H2O + 2 e-

**Step 9**

Before we can add the two half reactions to obtain the overall net ionic reaction, the number of electrons lost must equal the number of electrons gained.

Multiply each 1/2 reaction to equal the LEST COMMON MULTIPLE. In this case, the L.C.M. is 6

2x [ 3 e- + 2 H2O + (-)MnO4 --> MnO2 + 4 (-)OH ]

= 6 e- + 4 H2O + 2 (-)MnO4 --> 2 MnO2 + 8 (-)OH

3x [ 2 (-)OH + (2-)SO3 --> (2-)SO4 + H2O + 2 e- ]

= 6 (-)OH + 3 (2-)SO3 --> 3 (2-)SO4 + 3 H2O + 6 e-

**Step 10**

Add the two half reactions together, cancelling any like terms (6 e- on LEFT cancel with the 6 e- on right, etc) to obtain the balanced, net ionic reaction.

H2O + 2 (-)MnO4 + 3 (2-)SO3 --> 2 MnO2 + 2 (-)OH + 3 (2-)SO4

***FINAL STEP***

Check for mass and charge balance. You can add the K and Na back in if your instructor prefers.

H2O + 2 KMnO4 + 3 Na2SO3 --> 2 MnO2 + 2 KOH + 3 Na2SO4