Balancing Redox Equations: Half-reaction?

[marco9905]

I am having some problem with this question, which is:

Basic solution

Potassium permanganate, KMnO₄, is a powerful oxidizing agent. The products of a given redox reaction with the permanganate ion depend on the reaction conditions used. In basic solution, the following equation represents the reaction of this ion with a solution containing sodium sulfite:

MnO₄^-(aq) + SO₃^2-(aq) → MnO₂(s) + SO₄^2-(aq)

Since this reaction takes place in basic solution, H₂O(l) and OH^- (aq) will be shown in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation:

MnO₄^-(aq) + SO₃^2-(aq) + ______ → MnO₂(s) + SO₄^2-(aq) + _______

What are the coefficients of the six species in the balanced equation above? Remember to include H₂O(l) and OH^-(aq) in the blanks where appropriate

Here is what i understand from this question:

I know that you have to make 2 half reactions which ive done:
MnO₄^-(aq) → MnO₂(s)
4SO₃^2-(aq) → SO₄^2+(aq)

From my understanding i have to add H₂₀ to the first half reaction
MnO₄^-(aq) → MnO₂(s) +H₂O

This is where i get confused, i don't understand if i am meant to add H+ or OH-. If it is OH- how do you balance this half equation.

Also am i right in saying you would leave the 2nd half equation as it is other then adding electrons.

Hope all that makes sense and thanks in advance.

[Zedekiah]

Yep, going good. Actually, you would have to balance hydrogens for that half reaction. Half reactions have to be balanced in everything except electrons.

There are two methods to continue where you left off. One involves simply balancing with hydrogen ions and later adding an equal amount of hydroxide to either side to eliminate all of the hydrogen ions and form water.

I'll solve the problem step by step so you can see what I'm trying to say.

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MnO4(-) → MnO2

Add water to balance oxygen.

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MnO4(-) → MnO2 +2 H2O

Add hydrogen to balance hydrogen.

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4H(+)  + MnO4(-) → MnO2 +2 H2O

Add electrons to balance charge.

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4H(+) + 3e-  + MnO4(-) → MnO2 +2 H2O

Doing the same thing to this half reaction :

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SO3(2-) + H2O → SO4(2-) + 2 H(+)  + 2e-

Now, balance the electrons.

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2(4 H(+) + 3e-  + MnO4(-) → MnO2 +2 H2O)

3(SO3(2-) + H2O → SO4(2-) + 2 H(+)  + 2e-)

2 H(+) + 3 SO3(2-) + 2 MnO4(-) → 2 MnO2 + 3 SO4(2-) + H2O

Now, add enough hydroxide to each side to eliminate all of the acid.

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

Now,

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H(+) + OH(-) → H2O

and

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H2O

's on either side cancel to give :

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