Question

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Answer 1

Explanation:

a) The presence of sulfate ions in a solution can be confirmed by the reaction of barium chloride in an acidic medium.

The balanced chemical equation of the reaction is shown below:

$BaCl_2(aq)+CuSO_4(aq)->CuCl_2(aq)+BaSO_4(s)$

Hence, the white precipitate is barium sulfate and its formation with the ionic equation is shown below:

$Ba^2^+(aq)+SO_4^2^-(aq)->BaSO_4(aq)$

b) The presence of copper (II) ions can be confirmed by the following test:

Add potassium iodide solution to copper (II) solution.

Then a white ppt of cuprous iodide along with the liberation of iodine is observed and the entire solution attains brown color.

The chemical equation of the reaction is shown below:

$2CuSO_4(aq)+4KI(aq)->Cu_2I_2(s)+I_2(s)+2K_2SO_4(aq)$

c)(i)Due to this reaction, the blue color of the solution becomes white.

Reddish-brown copper is deposited at the bottom of the container.

(ii)In this reaction, zinc is oxidized.

d) (i) Copper is produced at the cathode.

(ii)$Cu^2^+(aq)+2e^-->Cu(s)$

(iii) The reaction that takes place at the cathode is reduction.

Reduction is gaining of electrons.