Question

Which of the following statements is true about a galvanic cell consisting of Cu and \( \mathrm{H_2} \) electrodes ?

[Given : \( \mathrm{E^{\circ}_{Cu^{2+}/Cu}} = +0.34\ \mathrm{V} \), \( \mathrm{E^{\circ}_{H^+/H_2,\,Pt}} = +0.00\ \mathrm{V} \)]

• A. Oxidation occurs at copper electrode
• B. Reduction occurs at \( \mathrm{H_2} \) electrode
• C. \( \mathrm{H_2} \) is cathode and Cu is anode
• D. \( \mathrm{H_2} \) is anode and Cu is cathode

Hint:

The electrode with the larger reduction potential (copper, 0.34 V) is the cathode, so hydrogen is the anode.

Answer 1

The Correct Option is D

Step 1: Compare the electrode potentials:
The standard reduction potential of \( \mathrm{Cu^{2+}/Cu} \) is +0.34 V.
The standard reduction potential of \( \mathrm{H^+/H_2} \) is 0.00 V.
The electrode with the higher reduction potential is reduced more easily.

Step 2: Assign the electrodes:
Copper has the higher value, so reduction happens at the copper electrode. Copper is the cathode.
Hydrogen has the lower value, so oxidation happens at the hydrogen electrode. It is the anode.
The cell is written as \( \mathrm{Pt(s) | H_2(g) | H^+(aq) || Cu^{2+}(aq) | Cu(s)} \).

Step 3: Write the reactions:
Anode: \( \mathrm{H_2 \rightarrow 2H^+ + 2e^-} \)
Cathode: \( \mathrm{Cu^{2+} + 2e^- \rightarrow Cu} \)
\[ \mathrm{E^{\circ}_{cell}} = 0.34 - 0.00 = +0.34\ \mathrm{V} \]

Step 4: Check each option:
(A) is false because copper is reduced, not oxidised.
(B) is false because hydrogen is oxidised.
(C) is false because the roles are reversed.
(D) is true.

Final Answer:
Hydrogen is the anode and copper is the cathode, option (D). \[ \boxed{\text{(D)}} \]