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# The cell in which the following reaction occurs : 2Fe^3+( aq) +2I^-( aq) 2Fe^2+ (aq)+ I (s) ® has E 0 cell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction

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.

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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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