Calculate the standard cell potentials of galvanic cells in which the following reactons take place :

(i) 2 Cr (s) + 3 Cd²⁺ (aq) → 2Cr³⁺ (aq) + 3 Cd(s)

(ii) Fe²⁺ (aq) + Ag+ (aq) → Fe³⁺ (aq) + Ag(s)

Given EºCr³⁺, Cr = –0.74 V, EºCd²+/ Cd = 0.40 V, EºAg+/Ag = 0.80 V, Eº Fe³⁺/ Fe²⁺ = 0.77 V        

Also calculate ∆rGº and equilibrium constants of the reactions.                   

(i)  Eºcell = Eºcathode – Eºanode = – 0.40 V – (–0.74 V) = +0.34 V,     

DrGº = – n FEºcell = – 6 mol × 96500 C mol^–1 × 0.34 V, = – 196860 CV mol^–1 = – 196860 J mol^–1 = – 196.86 kJ mol^–1,  – ∆rGº = 2.303 RT log K

196860 = 2.303 × 8.314 × 298 log K or log K = 34.5014, K = Antilog 34.5014 = 3.192 × 1034

(ii) Eºcell = + 0.80 V – 0.77 V = + 0.03 V.,                               

∆rGº = – n F Eºcell = – (1 mol) ´ (96500 C mol^–1) × (0.03 V),       

= – 2895 CV mol^–1 = – 2895 J mol^–1 = – 2.895 kJ mol^–1, – ∆rGº  = – 2.303 RT log K –2895 = –2.303 × 8.314 ´ 298 × log K , or log K = 0.5074 or K = Antilog (0.5074) = 3.22.