Find out the standard Gibbs free energy and the equilibrium constant o

1.	Find out the standard Gibbs free energy and the equilibrium constant of the cell reaction given below which has standard electrode potential of 0.236 V at 298 K.(Take F=96487 C, Antilog 0.981=9.37)2 Fe3+(aq) + 2I-(aq) → 2 Fe+2(aq) + I2(s)
Answer» We have given, E⁰ = 0.236 v at 298 k f = 96487 c 2 Fe³⁺(aq) + 2I^-(aq) → 2 Fe⁺²(aq) + I₂(s) We have, \(\triangle G^{\circ}=-nFE^{\circ}\) Here, n = 2 = No. of transfered electrons. So, \(\triangle G^{\circ}=-2 \times 96487 c\times0.236 v\) \(\triangle G^{\circ}=-45541.864 J\) At equilibrium \(\triangle G^{\circ}=-RTlnK_{eq}\) -45541.864 = -8.314 x 298 ln K_eq or 45541.864 = 8.314 x 298 x 2.303 x log K_eq ⇒ K_eq = 9.6 x 10⁷ Hence, for given reaction at 298k, standard cribbs free energy (\(\Delta G^{\circ}\)) = -4454.864 J and equilibrium constant = 9.6 x 10⁷

Find out the standard Gibbs free energy and the equilibrium constant of the cell reaction given below which has standard electrode potential of 0.236 V at 298 K.(Take F=96487 C, Antilog 0.981=9.37)2 Fe3+(aq) + 2I-(aq) → 2 Fe+2(aq) + I2(s)

Answer»

We have given, E⁰ = 0.236 v at 298 k

f = 96487 c

2 Fe³⁺(aq) + 2I^-(aq) → 2 Fe⁺²(aq) + I₂(s)

We have,

\(\triangle G^{\circ}=-nFE^{\circ}\)

Here, n = 2 = No. of transfered electrons.

So,

\(\triangle G^{\circ}=-2 \times 96487 c\times0.236 v\)

\(\triangle G^{\circ}=-45541.864 J\)

At equilibrium

\(\triangle G^{\circ}=-RTlnK_{eq}\)

-45541.864 = -8.314 x 298 ln K_eq

or 45541.864 = 8.314 x 298 x 2.303 x log K_eq

⇒ K_eq = 9.6 x 10⁷

Hence, for given reaction at 298k, standard cribbs free energy

(\(\Delta G^{\circ}\)) = -4454.864 J and

equilibrium constant = 9.6 x 10⁷

Discussion

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