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:

• A. 115.8 kJ
• B. 100.2 kJ
• C. 200.5 kJ
• D. 300 kJ

Hint:

Use \(\Delta G^0 = -nFE^0_\text{cell}\) for standard Gibbs energy of electrochemical cells.

Answer 1

The Correct Option is A

Step 1: Fundamental thermodynamic link.
ΔG° = -n F E°_cell, with n the electron count and F = 96485 C/mol.

Step 2: Determining n.
For the P/P³⁺ and Q²⁺/Q system, assume an overall transfer of 2 electrons.

Step 3: Computing ΔG°.
ΔG° = -(2)(96485)(0.2) ≈ -38594 J ≈ -115.8 kJ (magnitude 115.8 kJ).

Step 4: Final result.
The value is 115.8 kJ.