[MPQC]

Well. This question was originally discussed here, though I still don't understand how they came to the answer they discussed.

So basically, I'm trying to balance the equation. The full given equation is:

SO₃^2- + MnO₄^- + H⁺ Mn²⁺ + SO₄^2- + H₂O

So, I assigned each compound their respective oxidation numbers, and found that the following two elements changed:

SO₃^2- SO₄^2-

and

Mn Mn²⁺ + 2e^-

So what I don't get at this part, is how to balance them.

In the thread I mentioned above, they did the following to SO₃^2- to balance it:

SO₃^2- + H₂O S^2- + 2H⁺ + 2e^-

What I don't get is why they reversed which side H₂O and H⁺ is on. How am I supposed to know which elements go where, when they can just be reversed?

[Borek]

So, I assigned each compound their respective oxidation numbers, and found that the following two elements changed:

WHile it helps to find out what was reduced and what was oxidized, calculating ON is not necessary in the half cell method.

In the thread I mentioned above, they did the following to SO₃^2- to balance it:

SO₃^2- + H₂O S^2- + 2H⁺ + 2e^-

This is completely different reaction, first one was oxidation of sufurous ion to sulfate, this is reduction of sulfurous to sulfide. Are you sure it was presented this way? Can you link to the original thread?

[renge ishyo]

Boreks right, make sure you are careful to write your equations carefully so that they match the original equation. This might be partly responsible for the incorrect answers. For instance, you wrote this as a starting point:

Mn   Mn²⁺ + 2e-

You should have written this for your starting point (to match the original equation):

MnO₄^-   Mn²⁺

The next step is to balance the oxygens on each side, then the hydrogens, and then you balance the electrons. Hence, the +2e^- shouldn't be there either at this point.

SO₃^2- + H₂O    S^2- + 2H⁺ + 2e^-

What I don't get is why they reversed which side H2O and H+ is on. How am I supposed to know which elements go where, when they can just be reversed?

Again, look at the equation as there is a problem there on the right side...

To solve these problems, it is best to do things systematically step by step to build towards the answer. Here is an example for the first half reaction to get you started (try to do these same steps in the same order for the manganese half reaction):

1. Start by writing the half reaction.

SO₃^2-   SO₄^2-

2. Next, balance the sulfur atoms if necessary (it isn't in this example). Then balance the oxygens by adding H₂O molecules to the side that needs more oxygens. In the example below, it looks like this:

SO₃^2- + H₂O   SO₄^2-

3. Now there are 4 oxygens on each side, but now the hydrogens are unbalanced. Add H⁺ ions to the side of the equation that needs more hydrogens:

SO₃^2- + H₂O   SO₄^2- + 2H⁺

4. Now the final step is to make sure the net charge is the same on each side of the half reaction. On the left side of the equation we have a net charge of -2. On the right side, we have a net charge of O (the 2H⁺ and SO₄^2- cancel each other out). So to make the charges balance we need to add two extra negative charges to the right side of the equation.

SO₃^2- + H₂O   SO₄^2- + 2H⁺ + 2e^-

Now do the same thing for the manganese half reaction  

[khaly]

hiiiiiiiiiii,

SO32- + H2O= SO42- + 2H+ + 2e-) x5 => oxidation
MnO4- + 8H+ + 5e-= Mn2+ + 4 H2O)x2 => reduction
5 SO32- + 2 MnO4- + 6H+ = 5 SO42- + 2 Mn2+ + 3H2O


Cl2 + 2e-= 2 Cl- ) x5 ( reduction)
Cl2 + 6 H2O= 2 ClO3- + 12 H+ + 10e- ( oxidation)
6 Cl2 + 6 H2O = 10 Cl- + 2 ClO3- + 12 H+
3 Cl2 + 3 H2O =5 Cl- + ClO3- + 6H+

SO42- +8H+ + 8e-= S2- + 4 H2O ( reduction)
2 I- = I2 + 2e- ) x4 ( oxidation)
SO42- + 8I- + 8H+ = S2- + 4 I2 + 4 H2O

[AWK]

SO42- +8H+ + 8e-= S2- + 4 H2O ( reduction)
2 I- = I2 + 2e- ) x4 ( oxidation)
SO42- + 8I- + 8H+ = S2- + 4 I2 + 4 H2O

Impossible reaction

[Schrödinger]

SO42- +8H+ + 8e-= S2- + 4 H2O ( reduction)

SO₄^2- is just too stable to be reduced...atleast to S^2-. Not a common reaction that you can easily find.

[MPQC]

big wall of text

Thanks so much for the example! So here's what I got:

MnO₄^-   Mn²⁺

Then, to balance it..

MnO₄^- + 8H⁺ + 5e^- Mn²⁺ + 4H₂O

Then, I took both equations, and did as my text said, add them.

MnO₄^- + 8H⁺ + So₃^2- + H₂O + 5e^- Mn²⁺ + 4H₂O + SO₄^2- + 2H⁺ + 2e^-

And I'm pretty sure that it's correct, but I need to get rid of some of the extra elements. Can I just subtract them? For example, I have 1 mol of H₂O on the left side, and 4 moles of H₂O on the right side. Can I simply take one mole away from both sides, to have it equal to 0 on the left, and 3 on the right?

[Borek]

Yes, whatever is on both sides just cancels out.

[MPQC]

Thanks very much!

[renge ishyo]

Then, I took both equations, and did as my text said, add them.

MnO4- + 8H+ + So32- + H2O + 5e-   Mn2+ + 4H2O + SO42- + 2H+ + 2e-

You missed only one step...before adding the equations you have to make sure the electrons match on each half reaction so that they cancel out. Here is an example for this using your problem:

1. Start with the balanced half reactions.

SO₃^2- + H₂O    SO₄^2- + 2H⁺ + 2e^-

MnO₄^- + 8H⁺ + 5e^-    Mn²⁺ + 4H₂O

2. Multiply each equation so that the number of e^- match on each equation:

5 * (SO₃^2- + H₂O    SO₄^2- + 2H⁺ + 2e^-)

2 * (MnO₄^- + 8H⁺ + 5e^-    Mn²⁺ + 4H₂O)

To get:

5SO₃^2- + 5H₂O    5SO₄^2- + 10H⁺ + 10e^-

and

2MnO₄^- + 16H⁺ + 10e^-    2Mn²⁺ + 8H₂O

3. Now, add the two equations together. The electrons (e^-) should cancel at this point:

5SO₃^2- + 5H₂O + 2MnO₄^- + 16H⁺    5SO₄^2- + 10H⁺ + 2Mn²⁺ + 8H₂O

4. Now cancel other species such as H⁺ and H₂O on both sides as you correctly suggested:

5SO₃^2- + 2MnO₄^- + 6H⁺    5SO₄^2- + 2Mn²⁺ + 3H₂O

5. Finally, check the net charge on each side of the equation. If the two sides don't match you have made a mistake and need to go back. In the example above the net charge on the left side is 5(-2) + 2(-1) + 6(1) = -6. On the right side, the net charge is 5(-2) + 2(+2) = -6.

[Borek]

Wow, how I missed that... MnO₄^- + 8H⁺ seemed right and I have not looked further.

[renge ishyo]

Doesn't really matter, I think the student felt good about things and that is what is important. Gotta keep em going, ya know 

[MPQC]

Much thanks for the help. I corrected it . Finished a few more acid reactions nicely. Now..I have one last question...(I hope.)

The equation is:

Cl₂ + OH^- Cl^- + ClO₃^- + H₂O

So I progressed though the question nicely. It split into the following two parts:

Cl₂ Cl^- + ClO₃^-

And

OH^- H₂O

So I balanced the first one nicely, but I'm a tad confused on the second one. If I add H₂O on the left side, it'll just cancel out, and leave me with..Well..OH^-. The only way I can see an easy way of balancing it would be to add an H⁺ to the left side, but am I allowed to do that?

[renge ishyo]

This last reaction you have posted is for a basic solution. Therefore, "technically" you are supposed to use OH^- and H₂O to balance and not H⁺ at all (I say technically because I am far too lazy to do it that way, and so I about to show you a trick one of my favorite professors taught me ).

I will show you how to modify your method slightly to balance basic (OH^-) equations after doing the acidic (H⁺) balance steps to start off.

Start off with the two half reactions:

Cl₂   Cl^-

Cl₂   ClO₃^-

Note that OH^- or H⁺ and H₂O do not show up in the half reactions yet (just as H⁺ and H₂O did not show up in the half reactions of your first example). And yes, you can have one chemical species being oxidized and reduced in the same reaction (in the first equation chlorine is reduced, in the second it is oxidized).

2. Balance using H₂O, H⁺, and e^- as before (ignore OH^- for now).

2e^- + Cl₂   2 Cl^-

6H₂O + Cl₂   2ClO₃^- + 12 H⁺ + 10e^-

3. Multiply everything in the top equation by 5 so that the electrons in both equations cancel (should be 10e^- on both sides so that they cancel exactly), and then add the two equations together to get:

6 H₂O + 6Cl₂   2ClO₃^- + 10Cl^- + 12H⁺

Now, if the question had given this as an acidic solution (H⁺ and H₂O as the two species) you would be finished. But it gave you a basic solution (OH^- and H₂O as the two species), so you have to add in two more steps to convert your acidic answer to a basic one: 1) Neutralize the acid by adding OH^- to both sides of the equation. 2) Cancel waters. For the above example, here is how you do it:

Add OH^- to both sides in an amount to cancel the excess acid:

+ 12OH^- + 6H₂O + 6Cl₂   2ClO₃^- + 10Cl^- + 12H⁺ + 12OH^-

Neutralize the acid with the base on the right side to make water:

+ 12OH^- + 6H₂O + 6Cl₂   2ClO₃^- + 10Cl^- + 12H₂O

Cancel the waters to get the answer for a basic solution:

12OH^- + 6Cl₂   2ClO₃^- + 10Cl^- + 6H₂O

Finally check the charges on both sides to see that they match (-12 on the left, -12 on the right, should be good).

[MPQC]

Thanks very much, for the last time (for this topic!), you've really helped me a lot to understand this.