Question

The half reactions are:

1. \[\ce{Fe^3+ + e- -> Fe^2+}\]; E° = 0.76 V
2. \[\ce{Ag+ + e- -> Ag}\]; E° = 0.80 V

Calculate K_c for the following reaction at 25°C:

\[\ce{Ag+ + Fe^2+ -> Fe^3+ + Ag}\] (F = 96500 C mol⁻¹)

Answer 1

Given: The redox reaction \[\ce{Ag+ + Fe^2+ -> Fe^3+ + Ag}\]

And the half-cell reactions are

\[\ce{Fe^3+ + e- -> Fe^2+}\]; E° = 0.76 V

\[\ce{Ag+ + e- -> Ag}\]; E° = 0.80 V

n = 1

F = 96500 C/mol

\[\ce{E{^{\circ}_{cell}}}\] = 0.04 V

R = 8.314 J mol⁻¹ K⁻¹

T = 298 K

We know that 

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]

= 0.80 − 0.76

= 0.04 V

By using the relation between \[\ce{E{^{\circ}_{cell}}}\] and K_c

\[\ce{\Delta G^{\circ} = -nFE{^{\circ}_{cell}}}\]

ΔG° = −RT ln K_c

Equating both expressions:

\[\ce{nFE{^{\circ}_{cell}} = RT ln K_c​}\]

ln K_c = \[\frac{nFE{^{\circ}_{cell}}}{RT}\]

ln K_c = `(1 xx 96500 xx 0.04)/(8.314 xx 298)`

= `3860/2478.57`

ln K_c = 1.558

K_c = e^1.558

K_c = 4.75