Answer
$ \Delta G^{\circ} = 81 KJ $
$ K_{c} = 1.6 \times 10^{14} $
Work Step by Step
$Ce^{4+}(aq) +1e → Ce^{3+}(aq) $ $ E^{\circ} = +1.61V$
$Fe^{3+}(aq) + 1e → Fe^{2+}(aq)+ 1e $ $ E^{\circ} = +0.77 V$
$ \Delta G^{\circ} = -n F E^{\circ}_{cell} $
Where
$\Delta G^{\circ} = standard\ free\ energy\ change$
n = number of electrons involved in cell reaction= 1
F = Faraday’s constant= 96500 C/mol
$ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} $
= 1.61 V -0.77 V = 0.84 V
$ \Delta G^{\circ} = - 1 \times 96500 \times 0.84 V$ =- 81060 J = - 81 KJ
$ \Delta G^{\circ} = 81 KJ $
$ \Delta G^{\circ} = R \times T \times lnK_{c} $
$-81060 =
[- 8.314 \times 298\times2.303 \times logK_{c}]$
$K_{c} = $
$Antilog -81060\div[- 8.314 \times298 K\times2.303]$
$=1.6 \times 10^{14}$
$ K_{c} = 1.6 \times 10^{14} $
