Chemistry, asked by Anonymous, 1 month ago

Hello, help me. Please.

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Answered by prabinkumarbehera
1

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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