Use the standard potential at 298 K for the following redox couples: Use the sta

Question

Use the standard potential at 298 K for the following redox couples:

Use the standard potential at 298 K for the following redox couples Ag^+ (aq) + e^- rightarrow Ag^0 (s) E degree = +0.80 V Fe^2+ (aq) + 2e^- rightarrow Fe^0 (s) E degree = -0.44 V to calculate the standard potential of the cell Ag|AgNO_3(aq)||Fe(NO_3)_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_r H degree (Ag^+, aq) = 105.58 kJ mol^-1 and Delta H degree(Fe^2+, aq) = -89.1 kJ mol^-1

Explanation / Answer

The overall cell reaction is

Fe2+ (aq) + 2 Ag0 (s) ------> Fe0 (s) + 2 Ag+ (aq) …..(1)

The standard potential of the cell is

E0cell = E0red + E0ox = (-0.44 V) +(-0.80 V) = -1.24 V (ans)

The standard Gibb’s energy of the reaction is given by

G0 = -n*F*E0cell where n = 2 = number of moles of electrons transferred; F = 1 Faraday of electricity = 96485 C = 96485 J/V. Plug in values:

rG0 = -(2 mole)*(98485 J/V)*(-1.24 V) = -239282.8 J = (-239282 J)*(1 kJ/1000 J) = -239.2828 kJ (ans).

The standard enthalpy of formation of the ions is given in the table.

The enthalpy of the reaction is given by

rH0 = (2 mole)*fH0 (Ag+, aq) – (1 mole)*fH0 (Fe2+, aq) (the enthalpy of formation of pure elements is zero)

= (2 mole)*(105.58 kJ/mol) – (1 mole)*(-89.1 kJ/mol) = 300.26 kJ (ans)

Use the relation,

rG0 = rH0 – T*rS0 where T = (25 + 273) K = 298 K is the standard temperature and rS0 = standard entropy of the reaction. Plug in values.

(-239.2828 kJ) = (300.26 kJ) – (298 K)*rS0

===> rS0 = [(300.26 kJ) – (-239.2828 kJ)]/(298 K) = 1.810 kJ/K

Again use the above relation to find rG0 at 35C (T = 35 + 273 = 308 K). Assume rH0 and rS0 to be independent of temperature.

rG0 = rH0 – T*rS0 = (300.26 kJ) – (308 K)*(1.810 kJ/K) = -257.22 kJ (ans).

Use the relation rG0 = -n*F*E0cell to calculate E0cell at 35C. Plug in values and write

(-257.22 kJ) = (-2 mole)*(96485 J/V)* E0cell

===> E0cell = (-257.22 kJ)*(1000 J/1 kJ)/(-2 mole)(96485 J/V) = 1.33 V (ans).

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