Question:
The standard Gibbs energy \(\Delta G^0\) for the electrochemical cell:
\[
\text{P(s)} | \text{P}^{3+}(\text{aq}, 0.01\,\text{M}) || \text{Q}^{2+}(\text{aq}, 0.02\,\text{M}) | \text{Q(s)}
\]
with \(E^0_\text{cell} = 0.2 \, \text{V}\) is:
The standard Gibbs energy \(\Delta G^0\) for the electrochemical cell:
\[
\text{P(s)} | \text{P}^{3+}(\text{aq}, 0.01\,\text{M}) || \text{Q}^{2+}(\text{aq}, 0.02\,\text{M}) | \text{Q(s)}
\]
with \(E^0_\text{cell} = 0.2 \, \text{V}\) is:
Show Hint
Use \(\Delta G^0 = -nFE^0_\text{cell}\) for standard Gibbs energy of electrochemical cells.
Updated On: Jun 19, 2026
- 115.8 kJ
- 100.2 kJ
- 200.5 kJ
- 300 kJ
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use Gibbs free energy relation for electrochemical cell.
\[ \Delta G^0 = -n F E^0_\text{cell} \] where \(n\) = number of electrons transferred, \(F = 96485\, \text{C/mol}\).
Step 2: Identify \(n\).
For P\(^0\)/P\(^{3+}\) and Q\(^{2+}\)/Q, assume overall 2 electrons transferred.
Step 3: Calculate \(\Delta G^0\).
\[ \Delta G^0 = -(2)(96485)(0.2) \approx -38594\, \text{J} \approx 115.8\, \text{kJ} \]
Step 4: Final conclusion.
\[ \boxed{115.8\, \text{kJ}} \]
\[ \Delta G^0 = -n F E^0_\text{cell} \] where \(n\) = number of electrons transferred, \(F = 96485\, \text{C/mol}\).
Step 2: Identify \(n\).
For P\(^0\)/P\(^{3+}\) and Q\(^{2+}\)/Q, assume overall 2 electrons transferred.
Step 3: Calculate \(\Delta G^0\).
\[ \Delta G^0 = -(2)(96485)(0.2) \approx -38594\, \text{J} \approx 115.8\, \text{kJ} \]
Step 4: Final conclusion.
\[ \boxed{115.8\, \text{kJ}} \]
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