Hello, help me. Please.
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Answer:
2CuSO4 + 4HI → 2CuI + 2H2SO4 + I2
Explanation:
Chalcocyanite + Hydrogen Iodide = Kupra(I) Jodido + Sulfuric Acid + Diiodine
2 CuSO4 + 2 HI → Cu2SO4 + I2 + H2SO4
This is an oxidation-reduction (redox) reaction:
SVI + 2 e- → SIV
(reduction)
2 I-I - 2 e- → 2 I0
(oxidation)
CuSO4 is an oxidizing agent, HI is a reducing agent.
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