[aeromat2]

This must be solved using oxidation number method
H₂S + H₂O₂ --> S₈ + H₂O

I understand the following:
O : changes from -1 to -2 <-- gains e-, reduction, oxidizing agent
S : changes from -2 to 0 <-- loses e-, oxidation, reducing agent

I don't understand how I am going to balance this equation.

[DevaDevil]

okay, so you have 2 half cell reactions that make up this equation; one based on Sulfur, the other on Oxygen:

x H₂S --> y S₈ + z H⁺ + n e^-

and

u H₂O₂ + v H⁺ + m e^- --> w H₂O


can you balance those first? (the number of electrons transfered per atom S, or O is equal to the oxidation number change of that respectable element.)

then look at the electrons used in each half-cell and balance the whole accordingly.

[aeromat2]

This one also has to be solved out like this : http://www.saskschools.ca/curr_content/chem30_05/6_redox/redox1_4.htm

I can't split it into half-r'xns.

[DevaDevil]

the "half reactions" work in the same way:

x H₂S --> y S₈ + z H⁺ + n e^-

oxidation state of S goes from -2 to 0, as you indicated, so 2 electrons transfered per sulfur
this gives:
n = 2x (this is what you were required to do)

then you have 8 sulfur after the arrow, so:
x = 8y

then, for each H₂S, you have 2 protons after the arrow, so:
z = 2x

n = z = 2x = 16y

so the simplest form: y = 1, then x = 8 and n = z = 16

gives:
8 H₂S --> 1 S₈ + 16 H⁺ + 16 e^-



if you want to do this in the main reaction, it works similar, but then the oxidation state change is less defined.

[Vidya]

half reactions
oxidation :H2S -----> S
reduction: H2O2-----> H2O
Pick up one equation and follow the following steps
let us say reduction
H2O2-----> H2O
1.Balance atoms other than H (we will balance O here as it is undergoing a change in Oxidation number )
H2O2---> 2H2O
2.Add electrons on the required side (reduction so gain)
2e + H2O2----> 2H2O
3.Balance charge by adding H+ (in case of acidic medium and OH- in case of basic medium )
so I will add 2H+ on left side so that equation is balanced with respect to charges on both the sides
2e + 2H+  + H2O2 ------> 2H2O
4.Balance H and O by adding water
it is already balanced with respect to H and O
similarly balance the oxidation half reaction
add both the balanced half reactions after equalizing no of electrons gained with no of electrons lost
you can follow this method in balancing any redox reaction

[rabolisk]

It appears that the OP's teacher wants him to use the oxidation number method, which is ridiculous because it shouldn't matter how you arrive at the answer, only that it is right.