A metal spoon is to be electroplated with Silver. The electrolyte selected is sodium argentocyanide. Answer:
(i) What kind of salt is sodium argentocyanide?
(ii) Name the Cathode.
(iii) Name the Anode.
(iv) Give the reaction occuring at anode.
(v) Give the reaction occuring at cathode.
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Answer:
1.complex salt,
Complex salt when dissolved in an aqueous solution does not completely dissociate into ions because we know that it is a coordination compound and a coordination compound retains its identity in the aqueous solution. Therefore, sodium argentocyanide is a complex salt
2.Pure copper,
Pure copper acts as cathode.
3.
During electrolytic refining of copper, the anode is the positive electrode and is made from impure copper blocks. The electrolyte is a mixture of copper sulphate and sulphuric acid. The negative electrode is cathode.
4.
The reaction at the anode is an oxidation reaction that releases electrons. Those electrons then flow through a wire to the cathode, where a reduction reaction takes place. The reduction reaction occurs when the cathode gains additional electrons.
5.
The half-reaction on the cathode where reduction occurs is Cu2+ (aq) + 2e- = Cu(s). Here, the copper ions gain electrons and become solid copper. The entire reaction can be written by combining both half-reactions: Zn(s) + Cu2+ (aq) = Zn2+ (aq) + Cu(s).
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Verified answer
[tex]\green{\huge\underline{\sf{Answer}}}[/tex]
(1) Sodium argentocyanide is a complex salt.
(2) Sodium argentocyanide is preferred to silver nitrate as an electrolyte because the deposition of silver on the articles takes place slowly in the former, resulting in more smooth and uniform plating.
(3) One condition to ensure that the deposition is smooth, firm and long lasting is to make use of a low electric current passage (D.C. passage) for a long time during electroplating.
(4) Reaction at cathode: The silver ions gain electrons from the electrolyte to form silver atoms, that get deposited on the article as a layer of silver.
(fro melectrolyte) silver de positedo narticle
[tex]Ag ^{ + } + e ^ - { } \: -> Ag[/tex]
(5) Reaction at anode: The silver anode loses electrons into the electrolyte to become silver ions.
[tex]Ag→ Ag (in to electrolyte) + e ^{ - } [/tex]
[tex]\small\bf{\underline{\green{✔verified \: answer }}}[/tex]
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