Question

The overall formation constant for the reaction of 6 mole of $C{N}^{-}$ with cobalt (II) is $1\times {10}^{19}$. The standard reduction potential for the reaction ${\left[Co\right(CN{)}_6]}^{3-}$.
Given $C{o}^{3+}+e^{-}\to C{o}^{2+};E^{\circ }=1.82V$, if the ans can be expressed as $K_f=k\times {10}^{64}$. (write the value to the nearest integer)

Answer 1

Use $Co(CN{)}_6^{4-}\to Co(CN{)}_6^{3-}+e;E_{OP}^{\circ }=+0.83V$
$C{o}^{3+}+e\to C{o}^{2+};$ $E_{RP}^{\circ }=1.82V$.................................................................
$Co(CN{)}_6^{4-}+C{o}^{3+}\rightleftharpoons C{o}^{2+}+Co(CN{)}_6^{3-}$
and $E_{cell}=E_{cell}^{\circ }+\frac{0.059}{1}{\log }_{10}\frac{\left[C{o}^{3+}\right]\left[Co\right(CN{)}_6^{4-}]}{\left[C{o}^{2+}\right]\left[Co\right(CN{)}_6^{3-}]}$
or $E_{cell}=E_{cell}^{\circ }+\frac{0.059}{1}\log 10\frac{\left[C{o}^{3+}\right]\left[Co\right(CN{)}_6^{4-}\left]\right[C{N}^{-}{]}^6}{\left[C{o}^{2+}\right]\left[Co\right(CN{)}_6^{3-}\left]\right[C{N}^{-}{]}^6}$
Also $6C{N}^{-}+C{o}^{2+}\rightleftharpoons Co(CN{)}_6^{4-}$
and $K_{f_1}=\frac{\left[Co\right(CN{)}_6^{4-}]}{\left[C{o}^{2+}\right][C{N}^{-}{]}^6}$
and $6C{N}^{-}+C{o}^{3+}\rightleftharpoons Co(CN{)}_6^{3-}$
and $K_{f_2}=\frac{\left[Co\right(CN{)}_6^{3-}]}{\left[C{o}^{3+}\right][C{N}^{-}{]}^6}$
$\therefore E_{cell}=E_{cell}^{\circ }+\frac{0.059}{1}{\log }_{10}\frac{K_{f_1}}{K_{f_2}}$
$0=0.83+1.82+\frac{0.059}{1}{\log }_{10}\frac{{10}^{19}}{K_{f_2}}$
$\therefore \frac{K_{f_2}}{{10}^{19}}=8.23\times {10}^{44}$