Question:
The cell potential $E_{cell}$ for the following cell is:
\[
A(s) | A^+(aq, 0.1M) || B^{2+}(aq, 0.01M) | B(s)
\]
Given:
\[
E^\circ_{A^+/A} = 1V,\quad E^\circ_{B^{2+}/B} = 3V
\]
The cell potential $E_{cell}$ for the following cell is:
\[ A(s) | A^+(aq, 0.1M) || B^{2+}(aq, 0.01M) | B(s) \] Given: \[ E^\circ_{A^+/A} = 1V,\quad E^\circ_{B^{2+}/B} = 3V \]
\[ A(s) | A^+(aq, 0.1M) || B^{2+}(aq, 0.01M) | B(s) \] Given: \[ E^\circ_{A^+/A} = 1V,\quad E^\circ_{B^{2+}/B} = 3V \]
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Cathode has higher reduction potential; $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$.
Updated On: Jul 18, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Identify cathode and anode.
Higher reduction potential acts as cathode. Here: \[ E^\circ_{B^{2+}/B} = 3V \gt 1V \] So B is cathode and A is anode.
Step 2: Write standard cell potential.
\[ E^\circ_{cell} = 3 - 1 = 2V \]
Step 3: Apply Nernst correction idea.
Since concentrations are not standard, Nernst equation applies, but given MCQ typically assumes correction cancels or is negligible at this level unless explicitly required.
Step 4: Evaluate reaction direction.
Cell reaction favors electron flow from A to B, consistent with given potentials, so base value dominates.
Step 5: Final value.
\[ E_{cell} \approx 2V \]
Step 6: Final conclusion.
\[ \boxed{2.0} \]
Higher reduction potential acts as cathode. Here: \[ E^\circ_{B^{2+}/B} = 3V \gt 1V \] So B is cathode and A is anode.
Step 2: Write standard cell potential.
\[ E^\circ_{cell} = 3 - 1 = 2V \]
Step 3: Apply Nernst correction idea.
Since concentrations are not standard, Nernst equation applies, but given MCQ typically assumes correction cancels or is negligible at this level unless explicitly required.
Step 4: Evaluate reaction direction.
Cell reaction favors electron flow from A to B, consistent with given potentials, so base value dominates.
Step 5: Final value.
\[ E_{cell} \approx 2V \]
Step 6: Final conclusion.
\[ \boxed{2.0} \]
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