Chemistry Forum

Here are the two formulas:

Br- + KHSO₅ --> HOBr + H₂SO₄
Br- + O₃ + H₂O --> HOBr + O₂ + H+

Br- + KHSO₅ --> HOBr + H₂SO₄

Number of electrons from reducing agent (Br-) should be equal to number of electrons accepted by oxidizer (peroxide KOS(O₂)-O-OH): three oxygen atoms are O₂- and two (bold highlight) oxygen atoms of peroxide are 2O-
Br- - 2e- = Br+
2O- + 2e- = 2O₂-

Therefore, for every Br- we need two KHSO₅ containing two peroxide oxygen atoms O-. Write their correct ratio on the left side. Equalize by sulfur on the right (one on each side)

Br- + KHSO₅ --> HOBr + H₂SO₄
Potassium must end up with salt, if in the presence of H₂SO₄

look what else is missing and where (usually water). Anion on the left side should produce anion with the same number of charges on the right
Br- + KHSO₅ = HOBr + KSO₄- now it is equilibrated

It seems like another presentation of the same reaction could be correct too
Br- + KHSO₅ = OBr- + KHSO₄
but sulfuric acid is stronger that HBrO, and therefore the first equation is more scientifically correct

Try the second problem by yourself using electron equilibration technique first
Br- + O₃ + H₂O --> HOBr + O₂ + H+

Br- - 2e- = Br+
but this is more challenging:
3O- + e- => 3O₂- how many electrons should be added to the left side to make total 3x2 = 6 negative charges?
Answer: to three existing add three e-
3O- + 3e- = 3O₂- Now this is equilibrated

Satisfy the rule Number of electrons from reducing agent (Br-) should be equal to number of electrons accepted by oxidizer O-
Find common denominator and multiply:
3Br- - 6e- = 3Br+
6O- + 6e- = 6O₂-

The rest is easy! Remember, anion on the left and cation (proton) on the right are given to you on purpose for confusion. You should equilibrate and make equal charges on both sides.
Please give your answer here.
Good luck!