The proton, neutron, and electron are the three sub–atomic particles important in chemistry. These p — Physical Chemistry — Thermodynamics Chemistry Question
Redox potential, Gibbs free energy, and solubility
The proton, neutron, and electron are the three sub–atomic particles important in chemistry. These particles occupy two regions. Proton and neutron occupy the central place of the nucleus and electron the vast space outside the nucleus.
Neutron transfer does not take place in ordinary chemical reactions. Proton (hydrogen ion) transfer constitutes acid–base reactions. Electron transfer constitutes oxidation–reduction reactions. Oxidation–reduction reactions are essential for life. Photosynthesis and respiration are two prime examples. Oxidation–reduction reactions also allow key thermodynamics quantities to be measured as demonstrated in this problem.
The following information is given:
Ag+(aq) + e– → Ag(s) E° = 0.7996 V
AgBr(s) + e– → Ag(s) + Br–(aq) E° = 0.0713 V
∆fG°(NH3(aq)) = –26.50 kJ mol–1
∆fG°(Ag(NH3)2+(aq)) = –17.12 kJ mol–1
[VISUAL]
Calculate ∆fG°(Ag+(aq)).
Model Answer
Ag+(aq) + e– → Ag(s) E° = 0.7996 V
∆G° = ∆Gf°(Ag( s)) + ∆Gf°(e –) – ∆Gf°(Ag +(aq)) = – ∆fG°(Ag +(aq)) = – F ∆E°
Therefore, ∆fG°(Ag +(aq)) = F ∆E° = 77.15 kJ mol–1
Calculate the equilibrium constant for the following reaction at 25 °C:
Ag+(aq) + 2 NH3(aq) → Ag(NH3)2 +(aq)
Model Answer
Ag+(aq) + 2 NH3(aq) → Ag(NH3)2 +(aq)
∆G° = ∆fG°(Ag(NH3)2 +(aq)) – ∆fG°(Ag +(aq)) – 2 ∆fG°(NH3(aq)) =
= –17.12 kJ – 77.15 kJ – 2 (–26.50) kJ = – 41.27 kJ
ln Kf = –∆G° / RT = 16.65
Kf = [Ag(NH3)2+] / ([Ag+][NH3]^2) = e^16.65 = 1.7×10^7
Calculate the Ksp value of AgBr(s) at 25°C.
Model Answer
AgBr(s) → Ag+(aq) + Br–(aq) ∆E° = (0.0713 – 0.7996) V = – 0.7283 V
ln Ksp = n F ∆E° / R T = -28.347
Ksp = [Ag+] [Br–] = e–28.347 = 4.9×10–13
Calculate the solubility of AgBr in an aqueous solution of ammonia at 25°C.
c(NH3) = 0.100 mol dm–3
Model Answer
Let us assume: [Ag+] << [Ag(NH3)2+].
AgBr(s) → Ag+(aq) + Br–(aq) Ksp = 4.9×10–13
Ag+(aq) + 2 NH3(aq) → Ag(NH3)2 +(aq) Kf = 1.7×107
AgBr(s) + 2 NH3(aq) → Ag(NH3)2 +(aq) + Br–(aq) K = Ksp Kf = 8.31×10–6
Equilibrium concentrations:
[NH3] = 0.100 – 2S
[Ag(NH3)2+] = S
[Br–] = S
K = S^2 / (0.100 – 2S)^2 = 8.31×10^–6
S / (0.100 – 2S) = 2.88×10^–3
S = 2.9×10^–4 (mol dm^–3)
Check:
[Ag+] = Ksp / [Br–] = 1.7×10^–10 << [Ag(NH3)2+]
Thus, the solubility of AgBr is 2.9×10^–4 mol dm^–3
A galvanic cell using the standard hydrogen electrode as an anode is constructed in which the overall reaction is
Br2(l) + H2(g) + 2 H2O(l) → 2 Br–(aq) + 2 H3O +(aq).
Silver ions are added until AgBr precipitates at the cathode and [Ag+] reaches a concentration of 0.0600 mol dm–3. The cell voltage is then measured to be 1.721 V. Calculate ∆E° for the galvanic cell.
Model Answer
[Br–] = Ksp / [Ag+] = 4.89×10–13 / 0.0600 = 8.15×10–12
Using Nernst equation at 25 °C:
E = E° - (0.0592 / 2) * log([Br-]^2 * [H3O+]^2 / p(H2))
With standard hydrogen electrode, [H3O+] = 1 and p(H2) = 1 atm:
E = E° - (0.0592 / 2) * log((8.15×10^-12)^2)
E = E° - 0.0592 * log(8.15×10^-12)
1.721 = E° - 0.0592 * log(8.15×10^-12)
E° = 1.721 + 0.0592 * log(8.15×10^-12) = 1.065 V
Calculate the solubility of bromine in the form of Br2(aq) in water at 25°C.
Model Answer
In order to estimate the solubility of Br2(aq), we need to calculate the Gibbs free energy of the reaction:
Br2(l) → Br2(aq) ∆G° = ?
For the reaction:
Br2(l) + 2 e– → 2 Br–(aq)
E1° = 1.065 V, ∆G1° = –2 F E1° = –2.130 F V
Let us first calculate E2° for the half–cell reaction:
Br2(aq) + 2 e– → 2 Br–(aq), ∆G2° = –2 F E2°
From the Latimer diagram:
BrO3–(aq) + 6 H3O+(aq) + 6 e– → Br–(aq) + 9 H2O(l) E3° = 1.441 V
BrO3–(aq) + 5 H3O+(aq) + 4 e– → HOBr(aq) + 7 H2O(l) E4° = 1.491 V
2 HOBr(aq) + 2 H3O+(aq) + 2 e– → Br2(aq) + 4 H2O(l) E5° = 1.584 V
Then:
2 BrO3–(aq) + 12 H3O+(aq) + 10 e– → Br2(aq) + 18 H2O(l)
E6° = (2 × 4 E4° + 2 E5°) / 10 = 1.5096 V
Similarly, Br2(aq) + 2 e– → 2 Br–(aq)
E2° = (2 × 6 E3° – 10 E6°) / 2 = 1.098 V
(Note that 6 × E3° = 4 × E4° + 1 × E5° + 1 × E2°)
Then, ∆G2° = –2 F E2° = –2.196 F V
Finally, ∆G° = ∆G1° – ∆G2° = 0.066 F V = 6368 J mol−1
Therefore,
K = [Br2(aq)] = e^(-∆G° / RT) = e^(-2.569) = 0.077 mol dm-3