Question

Which of the following net cell reactions occurs in a galvanic cell containing cadmium electrode and standard hydrogen electrode?
\(\text{E}_{(\text{Cd}_{(aq)}^{2+}|\text{Cd}_{(s)})}^{\circ} = -0.403 \text{V}\).

• A. \(\text{H}_{2(g)}+\text{Cd}_{(aq)}^{2+}⟶2\text{H}_{(aq)}^++\text{Cd}_{(s)}\)
• B. \(\text{Cd}_{(s)}+2\text{H}_{(aq)}^+⟶\text{Cd}_{(aq)}^{2+}+\text{H}_{2(g)}\)
• C. \(2\text{H}_{2(g)}+\text{Cd}_{(aq)}^{2+}⟶4\text{H}_{(aq)}^++\text{Cd}_{(s)}\)
• D. \(2\text{Cd}_{(s)}+2\text{H}_{(aq)}^+⟶2\text{Cd}_{(aq)}^{2+}+\text{H}_{2(g)}\)

Hint:

The electrode with the more negative reduction potential acts as the anode and is oxidised.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a galvanic cell the electrode with the lower (more negative) reduction potential loses electrons and acts as the anode. The standard hydrogen electrode (SHE) has \(E^{\circ} = 0\) V by definition.

Step 2: Key Formula or Approach:
Compare \(E^{\circ}(\text{Cd}^{2+}|\text{Cd}) = -0.403\) V with \(E^{\circ}(\text{H}^+|\text{H}_2) = 0.000\) V. Then \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\) must be positive for a spontaneous reaction.

Step 3: Detailed Explanation:
Cadmium has the lower potential, so Cd is the anode: \(\text{Cd}(s) \to \text{Cd}^{2+} + 2e^-\).
Hydrogen ions are reduced at the cathode: \(2\text{H}^+ + 2e^- \to \text{H}_2(g)\).
Adding the two half reactions: \(\text{Cd}(s) + 2\text{H}^+(aq) \to \text{Cd}^{2+}(aq) + \text{H}_2(g)\).
\[ E^{\circ}_{cell} = 0 - (-0.403) = +0.403\ \text{V} \]
Option A is the reverse reaction, which would have \(E^{\circ} = -0.403\) V and would not occur. Options C and D are not balanced in charge or in atoms.

Final Answer:
The net reaction is \(\text{Cd}(s) + 2\text{H}^+ \to \text{Cd}^{2+} + \text{H}_2(g)\), option (B). \[ \boxed{\text{Cd}_{(s)}+2\text{H}^+_{(aq)} \to \text{Cd}^{2+}_{(aq)}+\text{H}_{2(g)}} \]