Answer to Question #161115 in Physical Chemistry for Cat

Question

The e.m.f. of the following cell reaction is 0.6753 V at 25 ℃ and 0.6915 V at 0 ℃. Cd(s) | CdCl2.5H2O(saturated) || AgCl(s) | Ag(s)

(i) Give the balance for the cell reaction

(ii) Calculate the free energy G, enthalpy H and entropy S, change of the cell reaction at 25 ℃.

Answer 1

here RHS is cathode and that is Agcl(s)|Ag(s) electrode

reduction half AgCl + e = Ag + Cl-

oxidation half Cd = Cd²⁺ + 2e

net cell reaction 2Agcl + Cd = Cd²⁺ + Ag + 2Cl^-1

G = -nFE = -2 x 96485 c x 0.6753 v = -130312.6 J

$\Delta$ S = (dE\dT) x nF = {(0.6753-0.6915)v\25k} x 2 x 96485 c = - (0.0162\25) x 192970 J K-1

= 125.04 Jk-1

$\Delta$ G = $\Delta$ H - T$\Delta$S

$\Delta$ H = $\Delta$G + T $\Delta$S = - 130312.6 J - 298 k x 125.04 Jk-1 = - 130312.6 J - 37261.92 J = - 167574.52 J