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# Find the cathode and anode product of electrolysis of ${\text{CuS}}{{\text{O}}_{\text{4}}}\left( {{\text{aq}}} \right){\text{/Cu}}$ electrode?

Last updated date: 13th Jun 2024
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Hint:Electrolysis is a process in which ionic compounds decompose into their elements by passing an electric current through the compound in liquid form. The anions get oxidized at the anode and in case of cations they are reduced at the cathode.

1) The electrolysis of ${\text{CuS}}{{\text{O}}_{\text{4}}}\left( {{\text{aq}}} \right)$ using a copper electrode (an active electrode) results in the transfer of copper metal from anode to cathode. Copper sulfate in aqueous solution ionizes to give:
${\text{CuS}}{{\text{O}}_{\text{4}}} \to {\text{C}}{{\text{u}}^{2 + }} + {\text{SO}}_4^{2 - }$
At cathode: ${\text{C}}{{\text{u}}^{2 + }}\left( {{\text{aq}}} \right) + 2{e^ - } \to {\text{Cu}}\left( {\text{s}} \right)$
At anode:${\text{Cu}}\left( {\text{s}} \right) \to {\text{C}}{{\text{u}}^{2 + }}\left( {{\text{aq}}} \right) + 2{e^ - }$
3) As we know that the reduction occurs at the cathode, then the positively charged copper ions will migrate towards the cathode where each ${\text{C}}{{\text{u}}^{2 + }}$ ion accepts two electrons to give copper metal which is deposited on the cathode. On the other hand, at the anode, each copper metal loses two electrons to give ${\text{C}}{{\text{u}}^{2 + }}$ ions.