Answer
$ E^{\circ}_{cell} = 0.37 V$
$ \Delta G^{\circ} = - 35.7 KJ $
$ K_{c} =1.99 \times 10^{6}$
Work Step by Step
$ Cu^{2+}(aq) +1e → Cu^{+}(aq) $ $ E^{\circ} = +0.15V$
$ Cu^{+}(aq) + 1e → Cu_{(s)} $ $ E^{\circ} = +0.52 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} $
= 0.52 V -0.15 V = 0.37 V
$ E^{\circ}_{cell} = 0.37 V$
$ \Delta G^{\circ} = - 1 \times 96500 \times 0.37 V$ =- 35705 J = - 35.7 KJ
$ \Delta G^{\circ} = R \times T \times lnK_{c} $
$-35705 = [- 8.314 \times 298\times2.303 \times logK_{c}]$
$K_{c} = Antilog [ -35705\div[- 8.314 \times298 K\times2.303]$
$ K_{c} =1.99 \times 10^{6}$
