Question

For a Galvanic cell consisting zinc electrode and standard hydrogen electrode,
\( \text{E}^\circ(\text{Zn}^{+2}_{(\text{aq}) \mid \text{Zn}_{(\text{s})}) = -0.76 \text{ V} \)}
Identify the reaction that takes place at positive electrode during working of cell?

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

Hint:

Anode = Negative (Oxidation); Cathode = Positive (Reduction) in a Galvanic cell.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A Galvanic (Voltaic) cell generates electrical energy from spontaneous redox reactions.
It consists of two half-cells. The electrode where oxidation occurs is the anode (negative terminal).
The electrode where reduction occurs is the cathode (positive terminal).
Electrons flow from the anode to the cathode through the external circuit.
Step 2: Key Formula or Approach:
1. Compare the standard reduction potentials (\( E^\circ \)) of the two electrodes.
2. The electrode with the higher (more positive) \( E^\circ \) has a greater tendency to be reduced and will act as the cathode (+).
3. The electrode with the lower (more negative) \( E^\circ \) has a greater tendency to be oxidized and will act as the anode (-).
Step 3: Detailed Explanation:
The given standard reduction potential for zinc is:
\( \text{E}^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76 \text{ V} \)
By convention, the standard reduction potential of the Standard Hydrogen Electrode (SHE) is exactly zero:
\( \text{E}^\circ(\text{H}^+/\text{H}_2) = 0.00 \text{ V} \)
Comparing the two potentials:
\( 0.00 \text{ V}>-0.76 \text{ V} \)
Since the SHE has a higher reduction potential, it will act as the cathode.
The Zinc electrode has a lower reduction potential, so it will act as the anode.
The question asks for the reaction at the positive electrode.
In a Galvanic cell, the cathode is the positive electrode.
At the cathode, reduction always occurs.
The reduction half-reaction at the hydrogen electrode is the gain of electrons by hydrogen ions:
\[ 2\text{H}^+_{(\text{aq})} + 2\text{e}^- \rightarrow \text{H}_{2(\text{g})} \]
(Note: Option C shows the oxidation of hydrogen, Option A shows the oxidation of zinc, and Option B shows the reduction of zinc. Only Option D shows the reduction of hydrogen.)
Step 4: Final Answer:
The reaction at the positive electrode is \( 2\text{H}^+_{(\text{g})} + 2\text{e}^- \rightarrow \text{H}_{2(\text{g})} \).