Question

Cr(SO4)3+Br2+KOH ➡️ K2CrO4+KBr+K2SO4+H2O

Answer 1

Сr2(SO4)3 + 3Br2 + 16KOH ----> 2K2CrO4 + 6KBr + 3K2SO4 + 8H2O
Br2(0) --(+2e)--> 2Br(-)        | 3  восстановление
2Cr(+3) --(-6e)--> 2Cr(+6)    | 1  окисление
Br2 - окислитель
Cr2(SO4)3 - восстановитель, за счёт Cr(+3)

Answer 2

The balanced equation for the reaction you provided is as follows:

2Cr(SO₄)₃ + 3Br₂ + 12KOH → 6K₂CrO₄ + 6KBr + 3K₂SO₄ + 6H₂O

This equation represents the reaction between chromium(III) sulfate (Cr(SO₄)₃), bromine (Br₂), and potassium hydroxide (KOH), which produces potassium chromate (K₂CrO₄), potassium bromide (KBr), potassium sulfate (K₂SO₄), and water (H₂O). Note that the coefficients in the balanced equation indicate the stoichiometric ratios between the reactants and products.