The following reaction provided the reaction equilibrium constant as K = 4.23 × 10 - 8 at 125°C. Fe2+(aq) + 2Ag(s) - Fe(s) + 2Ag*(aq) Calculate the cell potential at 125°C. Choose the closest one.

Question

0-0.13 V

0-0.58 V

0-0.29 V

0 -9.1 × 10 -2 v

0-1.25 V

0 -2.9 × 10 -3 v

0+1.16 V

Answer 1

Fe²⁺(aq) + 2Ag(s) ==> Fe(s) + 2Ag⁺(aq)

-nFEcell = -RT lnK

Ecell = RT lnK / nF

Ecell = (8.314 J/Kmol)(398K)(-16.98) / (2 mol)(96485 J/V-mol)

Ecell = -0.29 V