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Neon and Heteronuclear Diatomic Molecules

The overall f...

Question

The overall formation constant for the reaction of $6 m o l e$ of $C N^{-}$ with cobalt (II) is $1 \times 10^{19}$ . Given that
$C o (C N)_{6}^{3 -} + e \to C o (C N)_{6}^{4 -}; E_{R P}^{o} = - 0.83 V$
$C o^{3 +} + e \to C o^{2 +}; E_{R P}^{o} = 1.82 V$
If the formation constant for the reaction of $6 m o l e$ of $C N^{-}$ with cobalt (III) is $x \times 10^{63}$ then $100 x$ is________.

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Solution

Use $C o (C N)_{6}^{4 -} \to C o (C N)_{6}^{3 -} + e; E_{O P}^{o} = + 0.83 V$
$C o^{3 +} + e \to C o^{2 +}; E^{o} = + 1.82 V$
---------------------------------------------------
$C o (C N)_{6}^{4 -} + C o^{3 +} ⇌ C o^{2 +} + C o (C N)_{6}^{3 -}$
and $E_{c e l l} = E_{c e l l}^{o} + \frac{0.059}{1} l o g_{10} \frac{[C o^{3 +}] [C o (C N)_{6}^{4 -}]}{[C o^{2 +}] [C o (C N)_{6}^{3 -}]}$
or $E_{c e l l} = E_{c e l l}^{o} + \frac{0.059}{1} l o g_{10} \frac{[C o^{3 +}] [C o (C N)_{4}^{4 -}] [C N^{-}]^{6}}{[C o^{2 +}] [C o (C N)_{6}^{3 -}] [C N^{-}]^{6}}$
Also, $6 C N^{-} + C o^{2 +} ⇌ C o (C N)_{6}^{4 -}$
and $K_{f_{1}} = \frac{C o (C N)_{6}^{4 -}]}{[C o^{2 +}] [C N^{-}]^{6}} = 10^{19}$
and $6 C N^{-} + C o^{3 +} ⇌ C o (C N)_{6}^{3 -}$
and $K_{f_{2}} = \frac{[C o (C N)_{6}^{3 -}]}{[C o^{3 +}] [C N^{-}]^{6}}$
$∴ E_{c e l l} = E_{c e l l}^{o} + \frac{0.059}{1} l o g_{10} \frac{K_{f_{1}}}{K_{f_{2}}}$
$0 = 0.83 + 1.82 + \frac{0.059}{1} l o g_{10} \frac{10^{19}}{K_{f_{2}}}$ (At equilibrium $E_{c e l l} = 0$ )
$∴ \frac{K_{f_{2}}}{10^{19}} = 8.23 \times 10^{44}$
$∴ K_{f_{2}} = 8.23 \times 10^{63}$

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