Q.27 Given the standard reduction potentials (E⊖) for the half-cell reactions below, the standard Gibbs free energy of the dissolution of silver chloride in water, at 298 K, is _____________ J mol–1 (rounded off to nearest integer).
(Given: Faraday constant F = 96500 C mol–1; J = C × V)
AgCl(s) + e– → Ag(s) + Cl–(aq) ; E⊖ = 0.22 V at 298 K
Ag+(aq) + e– → Ag(s) ; E⊖= 0.80 V at 298 K
Calculate Standard Gibbs Free Energy of Dissolution of AgCl from Electrode Potentials
Final Numerical Answer
The standard Gibbs free energy of dissolution of AgCl(s) in water at 298 K is
+5.6 × 10⁴ J mol⁻¹ (≈ +55,870 J mol⁻¹, rounded to the nearest integer this is
55,870 J mol⁻¹).
Problem statement
The question provides two standard reduction half‑cell reactions at 298 K:
- AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq); E°₁ = 0.22 V
- Ag⁺(aq) + e⁻ → Ag(s); E°₂ = 0.80 V
The required overall dissolution reaction is:
AgCl(s) → Ag⁺(aq) + Cl⁻(aq)
Step 1: Construct the cell reaction
To get Ag⁺(aq) on the product side, reverse the second half‑reaction:
Ag(s) → Ag⁺(aq) + e⁻; E°₂ (reversed) = −0.80 V
Now add this to the first half‑reaction:
- First: AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq)
- Reversed second: Ag(s) → Ag⁺(aq) + e⁻
On adding, electrons and solid Ag cancel, giving the overall reaction:
AgCl(s) → Ag⁺(aq) + Cl⁻(aq)
The standard cell potential for this overall reaction is:
E°cell = 0.22 V + (−0.80 V) = −0.58 V
A negative E°cell already indicates that dissolution is non‑spontaneous under standard conditions.
Step 2: Relate E°cell to ΔG°
For any electrochemical reaction under standard conditions:
ΔG° = −n F E°cell
Where:
- n = number of electrons transferred = 1 (one electron in each half‑reaction)
- F = 96,500 C mol⁻¹ (Faraday constant; given)
- E°cell = −0.58 V
Substituting:
ΔG° = −(1)(96,500 C mol⁻¹)(−0.58 V)
Since 1 J = 1 C·V,
ΔG° = +55,970 J mol⁻¹ ≈ 5.6 × 10⁴ J mol⁻¹
Rounded to nearest integer (as asked):
ΔG° ≈ 55,870 J mol⁻¹
Many standard tables express this as about +55.6 kJ mol⁻¹, consistent with the very small Ksp of AgCl.
Thus, the dissolution of AgCl(s) in water under standard conditions is highly non‑spontaneous, reflected in the positive Gibbs free energy.
Concept explanation: why this method works
- Combining half‑reactions: Electrode potentials themselves do not add directly, but when half‑reactions are algebraically added, their corresponding Gibbs energies (and therefore nFE°) add, so the net E°cell found above is valid for the overall process.
- Sign of ΔG°: A positive ΔG° corresponds to K < 1 and hence very low solubility; in fact, the same data lead to Ksp ≈ 1.5 × 10⁻¹⁰ for AgCl at 298 K.
