Question

For the cell, Fe|Fe 2+ (0.01 M)|| Ag+ (0.1 M)|Ag, Write the cell reaction and calculate the emf of the cell at 298 K, if standard electrode potentials of Fe and Ag electrodes are -0.44 V and 0.8 V respectively. Calculate the change in standard free energy ∆Go for the reduction of 1 mole of Ag+.

Answer 1

Solution.

$EMF = Ec - Ea$

Since the standard electrode potential of silver is greater than the potential of iron, silver will be the cathode, and iron the anode.

$C: Ag^+ + 1e = Ag^0$

$A: Fe - 2e = Fe^{2+}$

General reaction scheme:

$Ag^+ + Fe = Ag^0 + Fe^{2+}$

$Ec = Eo + \frac{0.059*lg(Ag^+)}{1}$

$Ea = Eo + \frac{0.059 \times lg(Fe^{2+})}{2}$

Ec = 0.741 V

Ea = -0.499 V

EMF = 1.24 V

EMFo = 1.24 V

$\Delta G = -z\times F\times Eo$

z = 1

$\Delta G = -119641,4 \frac{kJ}{mole}$

Answer:

$C: Ag^+ + 1e = Ag^0$

$A: Fe - 2e = Fe^{2+}$

$Ag^+ + Fe = Ag^0 + Fe^{2+}$

EMF = 1.24 V

$\Delta G = -119641,4 \frac{kJ}{mole}$