Question
The overall formation constant for the reaction of 6 mole of CN− with cobalt (II) is 1 × 1019. What is the formation constant for the reaction of 6 moles of CN− with cobalt (III)?
Given that Co(CN)63− + e− → Co(CN)64−; E° = −0.83 V
Co3+ + e− → Co2+; E° = +1.81 V and 2.303 RT/F = 0.06.
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Solution
Given: $Co^{3 +} + e^{-} \rightarrow Co^{2 +};E^{o} = 1.81\text{ V;}\Delta G_{1}^{o} = - 1 \times F \times 1.81$
$Co(CN)_{6}^{3 +} + e^{-} \rightarrow Co(CN)_{6}^{4 -};E^{o} = - 0.83\text{ V;} $$${\Delta G_{2}^{o} = - 1 \times F \times ( - 0.83) }{Co^{2 +} + 6CN^{-} \rightarrow Co(CN)_{6}^{4 -};K_{f} = 10^{19}; }{\Delta G_{3}^{o} = - RT \times \ln 10^{19} }{\text{Required }Co^{3 +} + 6CN^{-} \rightarrow Co(CN)_{6}^{3 -};K_{f} = ?; }{\Delta G^{o} = - RT \times \ln K_{f} }{\text{Now, }\Delta G^{o} = \Delta G_{1}^{o} - \Delta G_{2}^{o} + \Delta G_{3}^{o} }{\text{Or, } - RT.\ln K_{f} = ( - 1.81F) - 0.83F + \left( - RT\ln 10^{19} \right) }{\text{Or, RT.ln}\frac{10^{19}}{K_{f}} = - 2.64F}$$
$\Rightarrow \log\frac{10^{19}}{K_{f}} = - \frac{2.64F}{2.303RT} = - \frac{2.64}{0.06} $$$\therefore K_{f} = 10^{63}$$
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