Question

Question #f7924

Answer 1

They react and form copper iodide(I) (`CuI`) and pottasium triodide(`KI_3`).
Thank you for Gaya3's post.

Explanation:

I thought that reaction with copper sulfate(`CuSO_4`) and potassium iodide(`KI`) does'nt happen for two reasons, but it was wrong.

This is what I have written:

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(1) Potassium cation (`K^+`) is very stable. `K` is one of alkali metals, which lies in the group 1 in the periodic table. In general,
alkali metal cations form no precipitation with any anions.

(2) Iodine anion (`I^-`) is also very stable. `I` is one of halogens,
in the group 17 in the periodic table. Halogen anions don't
precipitate expect with `Ag^+` and `Pb^(2+)`.

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What I did not notice was the fact that `I^-` has strong reducibility. Please see Gaya3's post for the chemical equation.

Answer 2

Yes!

Explanation:

Equation:

`color(blue)(CuSO_4) + color(red)(KI) -> color(orange)(CuI) + color(violet)(K_2SO_4) + color(green)(KI3`

After Balancing,

`color(blue)(2CuSO_4 + color(red)(5KI -> color(orange)(2CuI + color(violet)(2K_2SO_4 + color(green)(KI_3`

Watch it happen here