${2Fe}^{3 +}\ + \ {3I}^{-}\ \ \rightleftharpoons \ {2Fe}^{2 +}\ + \ I_{3}^{-}\ \ \ \ \$ The standard reduction potentials in acidic conditions are 0.77 and 0.54 V respectively for$\ {Fe}^{3 +}\ /\ {Fe}^{2 +}$ and $I_{3}^{-}\ \ /\ \ I^{-}$ couples. The equilibrium constant for the reaction is

Question

Dr. Abubacker · Accepted Answer

(c)
For the change , ${2Fe}^{3 +} + \ {3I}^{-}\  ightleftharpoons \ {2Fe}^{2 +}\  + \ I_{3}^{-}$
$E_{cell}^{{^\circ}}$ = 0.77 – 0.54 = 0.23 V
$E^{{^\circ}} = \ \frac{0.059}{2}\log K_{c}\$
$0.23\  = \ \frac{0.0591}{2}\log K_{c}\ \$
$K_{c}\  = \ 6.26\  	imes \ 10^{7}$

Q. ${2Fe}^{3 +}\ + \ {3I}^{-}\ \ \rightleftharpoons \ {2Fe}^{2 +}\ + \ I_{3}^{-}\ \ \ \ \$ The standard reduction potentials in acidic conditions are 0.77 and 0.54 V respectively for$\ {Fe}^{3 +}\ /\ {Fe}^{2 +}$ and $I_{3}^{-}\ \ /\ \ I^{-}$ couples. The equilibrium constant for the reaction is

Explanation

Reviewed by Dr. Abubacker

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