3.6    The cell in which the following reaction occurs:
             2Fe^{3+}(aq)+2I^{-}(aq)\rightarrow 2Fe^{2+}(aq)+I_{2}(s)    has E^{0}_{cell}=0.236\, V at 298 K.
            Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answers (1)

For finding Gibbs free energy we know the relation :- 

                                            \Delta G_r^{\circ} = -\ nFE_{cell}^{\circ}

                                                        = -\ 2\times96487\times 0.236

                                                        = -\ 45541.864\ J\ mol^{-1}

                                                        = -\ 45.54\ KJ\ mol^{-1}

Now, for equilibrium constant we will use :-

                                           \Delta G_r^{\circ} = -2.303\ RTlog\ K_c

So,                                        logK_c = -\frac{-45.54\times10^3}{2.303\times8.314\times298}

or                                          logK_c = 7.981   

or                                                K_c = 9.57\times10^7

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