I have a couple of redox reactions i just can't wrap my head around.
SO
2 + Br
2 HSO
4- + Br
-S on the left side has oxidation number +4 and on the right side +6. Br has the oxidation number 0 but reduces to -1. However, if i increase SO
2 to get the desired O
4 required on the right side i'm still left with an Extra S. Is it possible to equalize on the left side with H
3O
+ or is there some other way?
Cr
2O
72- + H
2S
Cr
3+ + S
8In this reaktion i found Cr's oxidation number to be +6 and S to be -2 on the left side and Cr to be +3 and S to be 0 on the right side. However i need at least 8 H
2S molecules to satisfy the right side and therefore need to multiply sulphur with 8 and 3 and Cr with 8 and 2. So 24 H
2S and 16Cr
2. However, on the left side i get a totalt oxidation number of 144, while's on the right side i only get 96. Should i just add more electrons on the left side or have i miscalculated somewhere?