
During the electrolysis of copper [II] sulfate solution using platinum as cathode and carbon as an anode. State the change noticed in the electrolyte.
a.) Blue color remain
b.) Blue color changes to red
c.) Blue color disappear
d.) None of these
Answer
561.9k+ views
Hint: Electrolysis: It is the process of separating a compound into its constituent elements by passing an electrical current through it when in a molten or aqueous state. The word electrolysis means the process of breaking molecules into smaller parts by using an electric current.
Complete Solution :
During the electrolysis of copper [II] sulfate solution using platinum as cathode and carbon as an anode. The following reaction will take place:
At cathode: ${ Cu }^{ +2 }{ +2e^{ - } }{ \rightarrow Cu }$
At anode: ${ 4OH }^{ - }{ +4e }^{ - }{ \rightarrow }{ O }_{ 2 }{ +H }_{ 2 }{ O }$
Copper metal is deposited at the cathode while oxygen gas and water at the anode.
As we know that ${ Cu }^{ +2 }$ ions are blue in color but they are getting discharged at the cathode, so the concentration of these ions in the solution decreases, so the blue color of the copper sulfate becomes light and the solution become colorless when all the copper ions get discharged at the cathode and the solution becomes colorless.
So, the correct answer is “Option C”.
Additional Information:
Characteristics of electrolysis:
1) On passing the electric current, cations migrate towards the cathode while the anions migrate towards the anode.
2) The number of electrons accepted by an anode is equal to the number of electrons donated by the cathode.
3) The preferential discharge of ions depends upon their position in the electrochemical series and the concentration of ions.
4) Neutral atoms of metals and hydrogen are liberated at the cathode. Therefore, they are called electropositive elements. Neutral atoms of non-metals are liberated at the anode. Therefore, they are called electronegative elements.
Note: The possibility to make a mistake is that the reduction takes place at the cathode while oxidation occurs at the anode in an electrolytic cell. Don’t confuse between these two.
Complete Solution :
During the electrolysis of copper [II] sulfate solution using platinum as cathode and carbon as an anode. The following reaction will take place:
At cathode: ${ Cu }^{ +2 }{ +2e^{ - } }{ \rightarrow Cu }$
At anode: ${ 4OH }^{ - }{ +4e }^{ - }{ \rightarrow }{ O }_{ 2 }{ +H }_{ 2 }{ O }$
Copper metal is deposited at the cathode while oxygen gas and water at the anode.
As we know that ${ Cu }^{ +2 }$ ions are blue in color but they are getting discharged at the cathode, so the concentration of these ions in the solution decreases, so the blue color of the copper sulfate becomes light and the solution become colorless when all the copper ions get discharged at the cathode and the solution becomes colorless.
So, the correct answer is “Option C”.
Additional Information:
Characteristics of electrolysis:
1) On passing the electric current, cations migrate towards the cathode while the anions migrate towards the anode.
2) The number of electrons accepted by an anode is equal to the number of electrons donated by the cathode.
3) The preferential discharge of ions depends upon their position in the electrochemical series and the concentration of ions.
4) Neutral atoms of metals and hydrogen are liberated at the cathode. Therefore, they are called electropositive elements. Neutral atoms of non-metals are liberated at the anode. Therefore, they are called electronegative elements.
Note: The possibility to make a mistake is that the reduction takes place at the cathode while oxidation occurs at the anode in an electrolytic cell. Don’t confuse between these two.
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