Question

Calculate the equilibrium constant for the reaction between fe2+(aq) and zn(s) under standard conditions at 25∘c.

Answer 1

Following reaction occurs in the given electrochemical system:
$Fe^{2+} + Zn$ → Fe + $Zn^{2+}$
Thus, under standard conditions
E(0) = E(0) Fe2+/Fe - E(0) Zn2+/Zn
where, $E^{0}Fe2+/Fe$ = standard reduction potential of Fe2+/Fe = -0.44 v
$E^{0}Zn2+/Zn$ = standard reduction potential of Zn2+/Zn = -0.763 v

E(0) = 0.323 v
now, we know that, ΔG(0) =-nFE(0) ............... (1)
Also, Δ$G^{0} = -RTln(K) ............ (2)$

On equating and rearranging equation 1 and 2, we get
K = exp( $\frac{nFE(0)}{RT}$ )= exp ($\frac{2X96500X0.323}{8.314X298}$) = 8.46 x $10^{10}$