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.
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Solution
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 Kc
`=>log K_c = (triangle_rG^theta)/(2.303 RT)`
`= (-45.54xx10^3)/2.303xx8.314xx 298`
= 7.981
∴Kc = Antilog (7.981)
= 9.57 × 107
Concept: Relation Between Gibbs Energy Change and Emf of a Cell
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