Login

calculate the emf of the given cell at 25C
Cd | Cd2 +(0.001M) || Fe2 +(0.6M) | Fe
standard electrode potential of Cd|Cd2+ and Fe|Fe2+ are -0.403V and -0.441 respectively

According to Nernst equation, 
Emf = E0 - 0.0591n log [Oxidised species][Reduced species]First Calculate E0E0 = EOx + ERed      = E0CdCd2+ + E0Fe2+Fe      = - 0.403 + - (-0.441)       = 0.038 VNow emf at 25 0C Emf = 0.038 - 0.05912 log [0.001][0.6]         = 0.038 - 0.05912 log 1.66 × 10-3        = 0.038 +0.0591 × 3 2 log 1.66           = 0.038 + 0.08865 × 0.2201            =0.0575 V

  • 1
View Full Answer
E°cell=E°cathode - E°anode
=-0.40-(-0.44)
=0.04 V

E°cell=log Kc×0.059/n
=>E°cell=logKc×0.059/2=0.04
=>logKc = 2×0.04/0.059= 1.356
=>Kc= antilog1.356=22.70
  • 7
What are you looking for?