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:
[tex]BaCl_2(aq)+CuSO_4(aq)->CuCl_2(aq)+BaSO_4(s)[/tex]
Hence, the white precipitate is barium sulfate and its formation with the ionic equation is shown below:
[tex]Ba^2^+(aq)+SO_4^2^-(aq)->BaSO_4(aq)[/tex]
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:
[tex]2CuSO_4(aq)+4KI(aq)->Cu_2I_2(s)+I_2(s)+2K_2SO_4(aq)\\[/tex]
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)[tex]Cu^2^+(aq)+2e^-->Cu(s)[/tex]
(iii) The reaction that takes place at the cathode is reduction.
Reduction is gaining of electrons.
