Question

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 3 I2(s) + 2 Fe(s) ? 2 Fe3+(aq) + 6 I-(aq) Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 3 I2(s) + 2 Fe(s) 2 Fe3+(aq) + 6 I-(aq)

3.5 × 10-59

8.9 × 10-18

1.7 × 1029

6.1 × 1058

1.1 × 1017

Please explain with each every detail explainations! Thanks

Answer 1

Fe³⁺ + 3?e^? --> Fe(s) Eo = -0.04 ---oxidation(anode)

I₂(s) + 2?e^? --> 2I- Eo = +0.54 ----- reduction(cathode)

Eocell = Ecathode - Eanode

= 0.54 - (-0.04) = 0.58V

dG = -nFEocell = -6 x 96487 x 0.58 (n=6 here because 6 electrons required to add during reaction.)

dG = -RTlnK = -8.314 x 298 x lnK

=> -8.314 x 298 x lnK =  -6 x 96485 x 0.58

K = 7.192 x 10^58

hence correct option is 4th 6.1 x 10^58 as i have taken here the standard potentials so there is small divergence with the answer you please do this with the given values of questions that you have not attach here surely you will get the correct solution.

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