Question

The cell in which the following reactions occurs: 2Fe³⁺ (aq) + 2I^– (aq) —> 2Fe²⁺ (aq) +I² (s) has E^°_cell=0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

 

Answer 1

Here, n = 2, `E_"cell"^theta` = 0.236  T = 298 K

We know that:

`triangle_rG^(theta) = -nFE_("cell")^theta`

= −2 × 96487 × 0.236

= −45541.864 J mol⁻¹

= −45.54 kJ mol⁻¹

Again, `triangle_rG^theta` = −2.303RT log K_c

`=>log K_c = (triangle_rG^theta)/(2.303 RT)`

`= (-45.54xx10^3)/2.303xx8.314xx 298`

= 7.981

∴K_c = Antilog (7.981)

= 9.57 × 10⁷