THE solubility of CuBr The EMF of the cell Pt|H2(g) (p =1.0 bar) | HBr(aq) (1.0×10−4 mol dm-3) | CuB — Physical Chemistry Chemistry Question

RH (Right Hand): CuBr(s) + e– → Cu(s) + Br−(aq)
LH (Left Hand): H⁺(aq) + e– → ½ H₂(g)

Nernst equation:
E = E° - (RT / F) * ln([H⁺][Br^-] / p(H₂)^(1/2))

Standard electrode potential for the CuBr electrode:
At 298 K, with p(H₂) = 1.0 bar and [H⁺] = [Br^-] = 1.0×10−4 mol dm^-3:
E = E° - (RT / F) * ln(1.0×10−8) = E° + 0.473 V
Since E = 0.559 V, E° = 0.559 V - 0.473 V = +0.086 V

Using ∆G° = −n * E° * F:
For the reduction of CuBr:
CuBr(s) + e^- → Cu(s) + Br^-(aq) (E° = +0.086 V)
∆G° = -1 * 0.086 V * 96485 C mol^-1 = −8.3 kJ mol−1

For the reduction of Cu⁺:
Cu⁺(aq) + e^- → Cu(s) (E° = +0.522 V)
∆G° = -1 * 0.522 V * 96485 C mol^-1 = −50.4 kJ mol−1

For the dissolution reaction (the difference of the two above reactions):
CuBr(s) → Cu⁺(aq) + Br^-(aq)
∆G° = -8.3 kJ mol^-1 - (-50.4 kJ mol^-1) = +42.1 kJ mol−1

Using ∆G° = −RT * ln(Ks):
Ks = exp(-∆G° / RT) = exp(-42100 / (8.3145 * 298.15)) = 4.2 × 10−8

Since [Br−(aq)] = 1.0×10−4 mol dm^-3 in the HBr solution,
[Cu⁺] = Ks / [Br^-] = 4.2 × 10−8 / (1.0 × 10−4) = 4.2 × 10−4 mol dm^-3

Using the Nernst equation:
E_new - E_old = (RT / F) * ln((p_new / p_old)^(1/2)) = (RT / ²F) * ln(2)
At 298 K:
∆E = (8.3145 * 298.15 / (2 * 96485)) * ln(2) = 0.0089 V (or a decrease of 8.9 mV if p(H₂) is on the reactant side of the net cell reaction)