Question

A galvanic cell consist copper electrode and standard hydrogen electrode. If \(E^{\circ}(\text{Cu}_{(aq)}^{+2}|\text{Cu}_{(s)}) = +0.34\,\text{V}\), then identify a reaction taking place at positive electrode during working of cell.

• A. \(\text{Cu}_{(s)}⟶\text{Cu}_{(aq)}^{+2}+2e^-\)
• B. \(\text{Cu}_{(aq)}^{+2}+2e^-⟶\text{Cu}_{(s)}\)
• C. \(\text{H}_2\text{(g)}⟶2\text{H}_{(aq)}^++2e^-\)
• D. \(\text{H}_{(aq)}^++2e^-⟶\text{H}_2\text{(g)}\)

Hint:

Cu2+/Cu has +0.34 V against SHE at 0 V, so Cu is the cathode (positive) and Cu2+ is reduced.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a galvanic cell the electrode with the higher reduction potential is the cathode. It is the positive electrode and reduction takes place there.

Step 2: Key Formula or Approach:
Compare the potentials. The standard hydrogen electrode (SHE) is defined as 0.00 V. The copper half-cell is +0.34 V.

Step 3: Detailed Explanation:
Since +0.34 V is greater than 0.00 V, copper has the greater tendency to be reduced, so the Cu electrode is the cathode and the positive terminal.
The reaction at the cathode is
\[ \text{Cu}^{2+}_{(aq)} + 2e^- \to \text{Cu}_{(s)} \]
The hydrogen electrode is the anode, where \(\text{H}_2 \to 2\text{H}^+ + 2e^-\) occurs (option C), and that is the negative electrode. Option (A) is oxidation of copper and (D) is reduction of \(\text{H}^+\), neither of which happens at the positive electrode in this cell.

Final Answer:
At the positive (copper) electrode the reaction is reduction of \(\text{Cu}^{2+}\), option (B). \[ \boxed{\text{Cu}^{2+}+2e^-\to\text{Cu} \text{ (B)}} \]