[funboy]

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

MnO₄^- --> Mn²⁺

5e^- + MnO₄^- + 8H⁺ --> Mn²⁺ + 4H₂O


SO₃^2- --> SO₄^2-

2H₂O + SO₃^2- --> SO₄^2- + 4H⁺ + 2e^-

For the MnO side of the equation to gain 5 electrons the SO4 side needs to loose 5 electrons
For the SO4 side to gain 2 electrons the MnO side of the equation needs to loose 2 electrons

Does this mean I need to do

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

5(2H₂O + SO₃^2- --> SO₄^2- + 4H⁺ + 2e^-)

??

Which then becomes

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

10H₂O + 5SO₃^2- --> 5SO₄^2- + 20H⁺ + 10e^-

and then added together and H2O and H+ canceled

2H₂O + 2MnO₄^- + 5SO₃^2- --> 2Mn²⁺ + 5SO₄^2- + 4H⁺

[funboy]

Im hoping this is correct

:/

If not, I think someone might need to hold my hand through this.

Chris

[chiralic]

check the CHARGE balance...

[funboy]

Ok, they are not equal

Left = 6+ right = 10+

help me find out where I went wrong??

[enahs]

Try balancing the charges in your half reactions.

[funboy]

quick question

when balancing the charges

is this correct

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

Mn = 7 (4 X -2 = -8 + 1 = -7)
0 = 2-
8H+ = 1

left side charge = 6+

Mn2+ = 2+
H= 1+
0 = 2-

right side charge = 1+

So I add 5e- to the left side

[chiralic]

Your 1st equation is OK... check the 2nd equation....

The charge balance in both side MUST BE equal....Also MUST BE equal THE NUMBER of elements in both side too

[funboy]

I must be getting tired because this is frustrating me

SO₃^2- --> SO₄^2-

S = 4+                        S = 6+
0 = 2-              ----->    O = 2-

There is a change of 2+ charge from left side to right side

so I would want to add 2e- to lower the charge on the right side correct??

so going through this step by step I would want to add 1 water to the side deficient in oxygen molecules then balance the hydrogen atoms by adding 1 H to the other side.

H₂O + SO₃^2- --> SO₄^2- + H⁺

Next I want to balance the electrical charge which I concluded was 2e- added to the right side of the equation


H₂O + SO₃^2- --> SO₄^2- + H⁺ + 2e^-

[Borek]

SO₃^2- --> SO₄^2-

S = 4+                        S = 6+
0 = 2-              ----->    O = 2-

There is a change of 2+ charge from left side to right side

so I would want to add 2e- to lower the charge on the right side correct??

You are mixing two things - half reaction method and ON method. When balancing by half reaction, don't look at ON, just balance atoms and charge as listed.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method

Your equation is at this moment already balanced with regard to charge - there is -2 on both sides, you don't have to do anything with that. Oxygen is not balanced, as there there is more oxygen on the right than on the left. Thus you have to balance oxygen with water and H⁺ - and once this is done balance charge:

H₂O + SO₃^2- --> SO₄^2- + H⁺ + 2e^-

Wrong, but good as a starting point. Remove all electrons. Balance hydrogen. Calculate total charge of molecules (don't touch ON, only charges are shown - +1 for H⁺ and -2 for both SO_x^2-) on both sides - and than add electrons to balance it. For obvious reasons you will have to add on the right number of electrons identical to number of H⁺ cations.

[funboy]

For the oddest reason Im having a difficult time calculating the charges for each side


I am guessing that the calculation of this only uses H+ and O charges

so left side = 2- for SO32-
right side = 2- for SO42- and 1+ for H

Does H2O come into play here H=1+ O=2-

For some reason I was under the impression that I could use difference in ON to calculate Half Cell equations, I guess it was a matter of luck that the charges worked on on my previous attempts (all but 2)

Let me know if Im on the right track

[funboy]

H2O + SO32- --> SO42- + 2H+ + 2e-

man Im making lots of small errors,

If someone can direct me to where I can learn to calculate the charges on the sides of reactions I would be greatful because its obvious I have no clue what I am doing in that respect (as well as many others)

[Borek]

Look ONLY at the charges listed for whole molecules.

H₂O - no charge at all (0)
H⁺        +1
SO₄^2-     -2
Fe³⁺     +3
ClO₄^-      -1

and so on. You are still trying to use ON (oxidation numbers) when balancing charges - don't do it! Doesn't matter what's inside of the ion, only thing that matters now is what charge does it have on the whole.

[funboy]

Awwww, looks so much easier

Damn brain, doesnt like to work on weekends


Thanks again

2 mins till I have to start my shift so I will redo(x) my reaction and post it, Im sure it will be correct this time

Chris