Question

A cell was prepared by dipping a Cu wire and a saturated Ag/AgCl electrode into 0.10 m CuSo₄ solution. The Cu wire was attached to the positive terminal of a potentiometer and the reference electrode was attached to the negative terminal. Write the half reaction for the Cu electrode

Answer 1

The half-reaction for the Cu electrode in the described galvanic cell setup is Cu(s) → Cu²⁺(aq) + 2e⁻, signifying the oxidation of solid copper to copper ions while releasing electrons that flow through the external circuit.

Step-by-step explanation:

The half-reaction for the copper electrode in a galvanic cell where a Cu wire is dipped into a 0.10 M CuSO₄ solution is the oxidation of copper. At the copper electrode, which is connected to the positive terminal of a potentiometer (suggesting it functions as the anode), the copper metal (Cu) loses two electrons to become copper ions (Cu²⁺). This process is represented by the half-reaction:

Cu(s) → Cu²⁺(aq) + 2e⁻

Here, Cu(s) stands for copper in its solid form, and Cu²⁺(aq) represents copper ions in aqueous solution. The electrons flow through the external circuit from the anode (Cu wire) to the cathode (in this case, a saturated Ag/AgCl electrode), where the reduction half-reaction involving silver ions takes place. Electrochemical cells like these convert the chemical energy of spontaneous redox reactions into electrical energy, allowing for the flow of electrons and thus electricity.