Question

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25

Answer 1

$3I_2+6e^-\rightarrow 6I^-$   

$2Fe\rightarrow 2Fe^3^++6e^-$ $E^o(oxidation/anode)=0.04V$

$E^o=E^o(reduction)+E^o(oxidation)$

$E^o=0.58V$

$E^o=\frac{RT}{nF}lnK$

$E^o=0.58V$

$R=8.314\frac{J}{molK}$

$T=298K$

$n=6e^-$

$F=96485\frac{Coulomb}{mole}$

$lnK=135.52$

$K=7.19*10^{58}$

$K\approx 2.4*10^5^8$

Answer 2

1.1

Answer 3

1.7