The standard cell potential for the following redox reaction is 0.236 V at 298 K : 2 Fe ³⁺( aq ) + 2 I ⁻( aq ) 2 Fe ²⁺( aq ) + I ₂( s ) The equilibrium constant ( K_ c ) for this reaction is: [ Given: 2.303RT F = 0.059 V at 298 K ]

Step-by-step solution

First, determine the number of electrons transferred ( n ) in the balanced redox reaction. The oxidation half-reaction is 2 I ⁻ I ₂ + 2 e ⁻ , and the reduction half-reaction is 2 Fe ³⁺ + 2 e ⁻ 2 Fe ²⁺ . Thus, n = 2 . The relationship between the standard cell potential and the equilibrium constant is: E^ _ cell = 0.059 n K_ c Substitute the given values ( E^ _ cell = 0.236 V and n = 2 ): 0.236 = 0.059 2 K_ c 0.236 = 0.0295 K_ c K_ c = 0.236 0.0295 = 8 K_ c = 10⁸ Answer: 1.0 10⁸