Use the standard potential at 298 K for the following redox couples Ag^+(aq) + e

Question

Use the standard potential at 298 K for the following redox couples Ag^+(aq) + e^- arrow Ag^0(s) E degree = +0.80 V Fe^2+(aq) + 2e^- arrow Fe^0(s) E degree = -0.44 V to calculate the standard potential of the cell Ag|AgNO3(aq)| | Fe(NO3)2(aq)|Fe and the standard Gibbs energy and enthalpy of the cell reaction at 25 degree C. Estimate the value of delta r G degree at 35 degree C. Calculate the standard potential of the cell at 35 degree C. The enthalpies of formation of Ag^+(aq) and Fe^2+(aq) are as given below. Delta f H degree (Ag^+, aq) = 105,58 kJ mol^-1 and delta f H degree (Fe^2+, aq) = -89.1 kJ mol^-1

Explanation / Answer

E0 = E0(Fe2+|Fe) - E0(Ag+|Ag) = -0.44-0.8 = -1.24V

Cell reaction: 2Ag + Fe2+ = 2Ag+ + Fe

No of electrons n = 2

F = 96500

At 25C or 298K

DG0(298) = -n*F*E0 = -2*96500*(-1.24) = 239320J = 239.32kJ/mol

At 25C or 298K

DG(298) = DHr - T*DSr

DHr = 2*DH(Ag+) - DH(Fe2+)

2*105.58 -(-89.1) = 300.26kJ/mol

DSr = (DHr - DGr(298))/298 =

(300.26-239.32)/298 =

0.204kJ/mol K

(i )

At 35 or 308K

DG(308) = DHr - T*DSr = 300.26 - 308*0.204 = 237.3kJ/mol = 237300J/mol

( ii )

At 35C or 308K

DG0(308) = -n*F*E0 =

E0(308) = -DG0(308)/(n*F) = -237300/(2*96500) = -1.23V

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