Question

Calculate the standard cell potentials of galvenic cells in which the following reactions take place:
(i) 2Cr_(s) + 3Cd²⁺_(aq) → 2Cr³⁺_(aq) + 3Cd_(s)
(ii) Fe²⁺_(aq) + Ag⁺_(aq) → Fe³⁺_(aq) + Ag_(s)

Given
E°_Cr³⁺/Cr = – 0.74 V, E°_Cd²⁺/Cd = 0.40 V
E°_Ag⁺/Ag = 0.80 V, E°_{Fe³⁺/Fe²⁺} = 0.77 V
Also calculate ArGo and equlibrium constants for the reactions.

Answer 1

(i) 2Cr _(s) + 3Cd_(aq)²⁺ → 2Cr³ + 3Cd_(s)
E°_cell = E°_Cathode – E°_Anode
= E°_(Cd²⁺cd) – E°_(Cr³⁺/Cr)
= – 0.40V – (-0.74V)
= – 0.40 V + 0.74V
= + 0.34V
∆G° = -nFE°_Cell
= – 6 mol x 96500 C mol^-1 x 0.34 V
= – 196860 CV mol^-1
= – 196.860 J mol^-1
∆G° = – 2.303RT log K_c
– 196.860 kJ = – 2.303 x 8.314 x 298 x logK_c

log K_c = 34.50/4
K_c= Antilog 34.50/4
= 3.173 x 10³⁴
Hence Gibb’s energy of cell (∆G°)= –196.86 kJ mol
Equilibrium constant (K_c) = 3.173 x 10³⁴
(ii) Fe²⁺_(aq) + Ag⁺_(aq) → Fe³⁺_(aq) + Ag_(s)
E°_Cell = E°_(Cathode) – E°_(Anode)
= E°_{(Ag⁺/ Ag)} – E°_{(Fe³⁺, Fe)}
= + 0.80 V – 0.77V
= 0.03V
∆G° = – n F E°_Cell
= – 1 mol^-1 x 96500 C mol^-1 x 0.03
= – 2895 J mol^-1
= – 2.895 kJ mol^-1
∆G° = – 2.303 RT log K_c
– 2895 = – 2.303 RT log K_c
– 2895 = – 2.303 x 8314 x 298 x logK_c

logK_c = 0.5074
K_c = Antilog 0.5074
K_c = 3.22
Gibb’s energy of cell = 2.895 kJ mol^-1
Equilibrium constant of cell = 3.22