Question

The cell in which the following reaction occurs:
2 Fe³⁺_(aq) + 2I^–_(aq) → 2 Fe²⁺_(aq) + I_2(s) has E°_Cell = 0.236 V at 298 K. Calculate the standard Gibb’s energy and the equilibrium constant for the cell reaction.

Answer 1

2 Fe³⁺ + 2e^– → 2 Fe ²⁺
2I^– → I₂ + 2e^–
Hence for the given cell reaction n = 2
∆_rG^o = – nFE°_Cell
= – 2 x 96500 x 0.236J
= – 45.55 kJ mol^-1
∆_rG^o = – 2.303 RT log Kc

= 7.983
K_c = Antilog (7.983)
= 9.616x 10⁷