Question

For the cell: Ag 's)| H2 (9) (Р %3D 1 bar), HCI (b = 1.00 molal), AgCl (s) | Ag(s) The standard cell potential was measured at different temperatures and the resulting data was fitted to a polynomial expression. The result is: E = 0.22233 - 0.0006477(T - 298.15) - 3.241 x 10-6(T - 298.15)2 Volts With T the Kelvin temperature What is the equilibrium constant for the cell reaction at 280 K?

Answer 1

The anode reaction is:

$\frac{1}{2}H_2(g)\rightarrow H^+(aq)+e$

and the cathode reaction is :

$AgCl(s) +e \rightarrow Ag(s)+Cl^-$

So, the reactions involve one electron transfer, so n=1.

Now, we know two equation of $\Delta G^0$

$\Delta G^0=-nFE^0$

$\Delta G^0=-RTln K$

$Thus, RT lnK = nFE^{0}$

$Thus, lnK = \frac{FE^0}{RT}$

Now according to the equation given, at T = 280 K,

E⁰ = 0.22233 - 0.0006477 x (280 - 298.15) - 3.241 x 10^-6x (280 - 298.15)² V = 0.233 V.

We know, F = 96485 C. g-eqv^-1, since here n = 1, so, F = 96485 C.mol^-1 , R = 8.314 J.K^-1.mol^-1 T = 280 K,

$Thus, lnK = \frac{96485C.mol^{-1}\times 0.233V}{8.314J.K^{-1}.mol^{-1}\times 280K} = 9.657$ (NOTE: C x V = J)

$Or, K = e^{9.657}=1.563 \times10^4$

This is the value of equilibrium constant.