Question

assuming that there was no overvoltage, what half reaction occurred at the anode of the electrolytic cell?

Answer 1

In an electrolytic cell, the half-reaction at the anode involves oxidation. For example, with a copper electrode at the anode, the reaction would be Cu(s) → Cu²⁺(aq) + 2e⁻, where copper loses electrons and is oxidized. The anode in an electrolytic cell is positively charged because the external voltage source drives the oxidation reaction.

Step-by-step explanation:

The half-reaction that occurs at the anode of an electrolytic cell involves oxidation, which is the loss of electrons. Since the question stipulates no overvoltage and provides the example of a copper electrode, the half-reaction can be represented as follows for the copper electrode:

• Cu(s) → Cu²⁺(aq) + 2e⁻

This oxidation reaction results in the copper metal losing electrons and thus, copper ions (Cu²⁺) are produced at the anode. In the context of an electrolytic cell, the anode is the electrode where oxidation occurs and it is also where positive ions migrate to. The external power supply drives the nonspontaneous reaction by applying a voltage that is higher than the reverse of the standard cell potential, making the copper electrode serve as the anode and causing it to be positively charged due to the oxidation.

Within a galvanic cell, which operates spontaneously, the anode would be negative, but in this case of an electrolytic cell, the anode is positive because it is where the external voltage source drives the oxidation reaction. The standard electrode potential is the potential for the reduction reaction under standard conditions, and it is reversed at the anode for oxidation to occur in an electrolytic cell.

Answer 2

In an electrolytic cell, the half-reaction at the anode involves oxidation. For example, with a copper electrode at the anode, the reaction would be Cu(s) → Cu²⁺(aq) + 2e⁻, where copper loses electrons and is oxidized. The anode in an electrolytic cell is positively charged because the external voltage source drives the oxidation reaction.

Step-by-step explanation:

The half-reaction that occurs at the anode of an electrolytic cell involves oxidation, which is the loss of electrons. Since the question stipulates no overvoltage and provides the example of a copper electrode, the half-reaction can be represented as follows for the copper electrode:

• Cu(s) → Cu²⁺(aq) + 2e⁻

This oxidation reaction results in the copper metal losing electrons and thus, copper ions (Cu²⁺) are produced at the anode. In the context of an electrolytic cell, the anode is the electrode where oxidation occurs and it is also where positive ions migrate to. The external power supply drives the nonspontaneous reaction by applying a voltage that is higher than the reverse of the standard cell potential, making the copper electrode serve as the anode and causing it to be positively charged due to the oxidation.

Within a galvanic cell, which operates spontaneously, the anode would be negative, but in this case of an electrolytic cell, the anode is positive because it is where the external voltage source drives the oxidation reaction. The standard electrode potential is the potential for the reduction reaction under standard conditions, and it is reversed at the anode for oxidation to occur in an electrolytic cell.