Given E_ Fe ³⁺ / Fe ²⁺ ^ =+0.76 ~V and E_ I ₂ / I ⁻ ^ =+0.55 ~V . The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is [ 2.303 R T F =0.06 ]

Step-by-step solution

E_ Fe ³⁺ / Fe ²⁺ ^ =0.76 ~V and E_ I ₂ / I ⁻ ^ =0.55 ~V K= ? Also 2.303 R T F =0.06 Here, I ₂ / I ⁻ will act as anode and Fe ³⁺ / Fe ²⁺ will act as cathode. 2 Fe ³⁺+2 I ⁻ 2 Fe ²⁺+ I ₂ Here, n=2 aligned E_ cell ^ &=E_ cathode -E_ anode =0.76-0.55=+0.21 G^ &=-n F E_ cell ^ =-2(96500) 0.21 &=-40530 ~J aligned Also, G=-2.303 R T K_ c aligned So, K_ c &= G - ( 2303 R T F ) F &= +40530 -(0.06) 96500 = 40530 5790 K_ c &=7 K_ c &=1 10⁷ aligned