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Given the overall formation constant of the $[Fe(CN)_6]^{-4}$ ion as $10^{35}$ and the standard potential for half reaction $Fe^{+3}+e^-\leftrightharpoons Fe^{+2},E^0=0.77V$,$[Fe(CN)_6]^{-3}+e^-\leftrightharpoons [Fe(CN)_6]^{-4},E^0=0.36V$.Then the overall formation constant of $[Fe(CN)_6]^{-3}$ ion is

$\begin{array}{1 1}(a)\;8\times 10^{92}\\(b)\;6.3\times 10^{14}\\(c)\;8.59\times 10^{41}\\(d)\;9\times 10^{36}\end{array}$

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Let $K_f$ be formation constant of $[Fe(CN)_6]^{-3}$ ion
$Fe^{+2}+6CN^-\leftrightharpoons [Fe(CN)_6]^{-4}$
$K_f=10^{35}$,$\Delta G_1^0=-2.303RT\log K_f$=-199704.6J
$Fe^{+3}+e^-\leftrightharpoons Fe^{+2}$
$\Delta G_2^0=-96500\times 0.77$
$\Rightarrow -74305J$
$Fe(CN)_6]^{-4}\leftrightharpoons [Fe(CN)_6]^{-3}+e^-$
$\Delta G_3^0=96500\times 0.33$
$\Rightarrow 34740J$
$Fe^{+3}+6CN^-\leftrightharpoons [Fe(CN)_6]^{-3}$
$\Delta G_4^0=\Delta G_1^{\large\circ}+\Delta G_2^{\large\circ}+\Delta G_3^{\large\circ}$
$\Delta G_4^0=-2392.69$Joule
$\Delta G_4^0=-2.303RT\log K_f$
$K_f=8.59\times 10^{41}$
Hence (c) is the correct answer.
answered Feb 18, 2014 by sreemathi.v

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