Question

Given:
\( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V} \)
\( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V} \)
What is the EMF of the galvanic cell:
\(\text{Zn} \mid \text{Zn}^{2+} \parallel \text{Cu}^{2+} \mid \text{Cu}\)?

• A. \(+1.10 \, \text{V}\)
• B. \(-1.10 \, \text{V}\)
• C. \(+0.42 \, \text{V}\)
• D. \(+0.76 \, \text{V}\)

Hint:

EMF of a galvanic cell = \( E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \). Remember to identify correctly which electrode is anode and cathode based on the standard potentials.

Answer 1

The Correct Option is A

Step 1: Identify the anode and cathode:
- Zinc electrode (\(\text{Zn}\)) undergoes oxidation: \[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \] - Copper electrode (\(\text{Cu}^{2+}\)) undergoes reduction: \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \] Step 2: Use standard electrode potentials:
\[ E^\circ_{\text{cathode}} = +0.34 \, \text{V} \] \[ E^\circ_{\text{anode}} = -0.76 \, \text{V} \] Step 3: Calculate the EMF of the cell: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 \, \text{V} \] Step 4: Therefore, the EMF of the galvanic cell is \( +1.10 \, \text{V} \).