Question

A voltaic cell consists of a Pb/Pb²⁺ half-cell and a Cu/Cu²⁺ half-cell at 25 ^oC. The initial concentrations of Pb²⁺ and Cu²⁺ are 0.0500 M and 1.50 M, respectively. What are the concentrations of Pb²⁺ and Cu²⁺ when the cell potential falls to 0.370 V?

Cu²⁺_(ag) + Pb_(s) → Cu_(s) + Pb²⁺_(ag)       

Hint: [Pb²⁺] + [Cu²⁺] = (1.5M + 0.05M) = 1.55 M total

Hint: the standard potential of Pb²⁺ + 2e^-   → Pb(s) is -0.130 and the standard potential of Cu²⁺ + 2e^-  → Cu(s) is +0.340.

The equation is:

E= E^o- (RT/nF) (LnQ)

Answer 1

22 - Cell Reaction - cü ing) cute + Pbce) → (46) + P3+(ay) Escue = E catode - Eanode Ecattede = Einth/2y = 0.34V Eanode a Epste/p6 = -0.1Bv Eocene = 0.34v - C-0.13) - 0:47V Condentsations Intions." ] = 0.05m 72 changa-x Cpyta] =1.som concertations cuth + Pb es) > eues + ph Lary) Inting oos loso +x. final (0.05-x) (1.50 +x) I at standard Conditions Nernst Egnation Esto a to = 0.470 R = gas coustat = 8.314 ] Temperature To 2984 na no of electson's ar F = 96,480 Colonna's Q = Quintout a CP32] a Ccute] 170K =298 kill

E= 0.37V 2- .. CV - CPT E - E 8.314 X218 2896480 Truth 0.01283 in - 50+] 70.05-x). 0.47 0.370 = In (1156+23 - 0.37V 0:47V -0.01283 Coos-se) oilo 0.01283 + In Coloso tx) = 7.754. x =e Como sage) 7.794 1.50th se 3 . 0.05-x (losotan = 2.426 X10 looose) loso tx = 2.426 X10 Co.os-x). 1o 50 tx = 12103 - 24262.. 2 +2426x = 12103 -1.50, 24272 - 11908. a= 11908 7127 2=000493

Final concertsation of U Crutz) = 0.05 cm = 0.08.-0.04g3 = 0.0007 Cpute] =oliste = 105 + 0.0493 = 1.5493