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Chemistry Redox equations help please.

ReTurd

qestion below.

i dont know why there is a picture of that man there, i cant get rid of it either.

(edited 11 years ago)

fdfgs.jpg3.2KB

Reply 1

arvin_infinity

Original post by ReTurd

Hello. im stuck on this question. [pics below]
Following the method used in the example i get an answer that is wrong because the h2o molecules dont balance, i got: 4HBr + H2SO4 -----> 2Br2 + SO2 + ??H2O

the final answer shows there only being 2 HBr and 1 Br2 molecules in the equation. the reason i doubled mine is because i followed the instuctions in the example, (third line on step 2) where is says to balance any atoms that change exidation number in your draft equation.

So why have they decided to use different methods in the two examples? It doesnt make sense.. how do i know if i should or shouldnt do that balancing thing??

Attachment not found

Attachments are not valid!

Reply 2

Fusionary

1) H2SO4 + 2e- + 2H+ --> SO2 + H2O
2) 2Br- --> Br2 + 2e-

Add them together to get:

3) H2SO4 + 2HBr --> SO2 + H2O + Br2

Reply 3

ReTurd

Original post by arvin_infinity

Attachments are not valid!

sorry, i think i got it working now

Reply 4

ReTurd

Hello. im stuck on this question. [pics below]
Following the method used in the example i get an answer that is wrong because the h2o molecules dont balance, i got: 4HBr + H2SO4 -----> 2Br2 + SO2 + ??H2O

the final answer shows there only being 2 HBr and 1 Br2 molecules in the equation. the reason i doubled mine is because i followed the instuctions in the example, (third line on step 2) where is says to balance any atoms that change exidation number in your draft

So why have they decided to use different methods in the two examples? It doesnt make sense.. how do i know if i should or shouldnt do that balancing thing?

2012-05-05 18.12.57.jpg

(edited 11 years ago)

Reply 5

Booyah

Why don't you just put a 2 in front of HBr and a 2 in front of H2O, correct me if I am wrong but that is balanced...

Reply 6

EierVonSatan

I can't see your images :colondollar:

Reply 7

ReTurd

Hello. im stuck on this question. [pics below]
Following the method used in the example i get an answer that is wrong because the h2o molecules dont balance, i got: 4HBr + H2SO4 -----> 2Br2 + SO2 + ??H2O

the final answer shows there only being 2 HBr and 1 Br2 molecules in the equation. the reason i doubled mine is because i followed the instuctions in the example, (third line on step 2) where is says to balance any atoms that change exidation number in your draft

So why have they decided to use different methods in the two examples? It doesnt make sense.. how do i know if i should or shouldnt do that balancing thing?

2012-05-05 18.12.57.jpg

Reply 8

ReTurd

Original post by EierVonSatan

I can't see your images :colondollar:

i think i have it fixed now ^^ in the post above. thanks

Reply 9

EierVonSatan

Original post by ReTurd

i think i have it fixed now ^^ in the post above. thanks

Cools

I'll just go through the question with the example method they've outlined (there are lots of methods to balance equations, you might find another easier).

Step 1: Write down what you're told

HBr + H2SO4 ---> Br2 + SO2

Step 2: oxidation changes

You are told to balance any atom that changes oxidation number so:

2HBr + H2SO4 ---> Br2 + SO2 (now S and Br are balanced, but H and O are not)

now compare the oxidation states

2HBr + H2SO4 ---> Br2 + SO2

Sulfur has gone from +6 to +4 (reduction = -2)

2HBr + H2SO4 ---> Br2 + SO2

bromine has gone from 2 x -1 to 2 x 0 (oxidation = +2)

Step 3: Balance oxidation changes

Nice and easy this time, they already balance :smile:

Step 4: Clean up

2HBr + H2SO4 ---> Br2 + SO2

two extra oxygens on the left, 4 extra hydrogens - add two waters to the right

2HBr + H2SO4 ---> Br2 + SO2 + 2H2O

That help?

why neg

Original post by EierVonSatan

Cools

Step 4: Clean up

2HBr + H2SO4 ---> Br2 + SO2

two extra oxygens on the left, 4 extra hydrogens - add two waters to the right

2HBr + H2SO4 ---> Br2 + SO2 + 2H2O

That help?

i can see what you have done there, but aren't you supposed to ignore the coefficient/number in front of the molecule when assigning oxidation numbers (this is why my answer is different and why i ended up getting an oxidation number of only -1 and ended up doubling it.

e.g. 2H2SO4 and H2SO4
Each H has an oxidation number of +1 in both, and the same with the other atoms. So in the 2H2SO4 molecule you cant say the the oxidation number of H is 2 x +1 = +2

does that make sense, am i right, wrong?

(edited 11 years ago)

Reply 12

EierVonSatan

Original post by ReTurd

It's quite subtle. It's not saying the oxidation number is +2 for hydrogen just that there are 2 hydrogens of oxidation state +1.

In the example above, it's saying that the overall oxidation change over the reaction has increased by +2 not that the individual oxidation state of each bromide is -2 on the left. In other words you need to lose 2 electrons to oxidise 2 bromide atoms - one for each bromide in the reaction.

These overall quantities in brackets is the bit relating to step 3.