Question

The overall formation constant for the reaction of 6 mol of CN^- with Cobalt (II) is 1 × 10¹⁹. The formation constant for the reaction of 6 mol of CN^- with Cobalt (III) is ______ × 10⁶³. Given that,

\[\ce{CO(CN)^{- 3}_6 + e^-1 -> Co(CN)^{-4}_6}\]

\[\ce{E^{\circ}_{R_p}}\] = 0.83 V

\[\ce{Co^{+3} + e^{- 1} -> Co^{+2}}\]

\[\ce{E^{\circ}_{R_p}}\] = 1.82 V

Options

• 5.63

• 2.30

• 8.23

• 9.36

Answer 1

The overall formation constant for the reaction of 6 mol of CN^- with Cobalt (II) is 1 × 10¹⁹. The formation constant for the reaction of 6 mol of CN^- with Cobalt (III) is 8.23 × 10⁶³.

Explanation:

\[\ce{[CO (CN)_6]^{4-} -> [CO (CN)_6]^{3-} + e^-}\]

\[\ce{CO^{3+} + e -> CO^{2+}}\]

\[\ce{[CO (CN)_6]^{4-} + CO^{3+} <=> CO^{2+} + [CO (CN)_6]^{3-}}\]

E_cell = `"E"_"cell"^circ + 0.059/1 log_10  (["Co"^(3+)]["Co" ("CN")_6]^(4-))/(["Co"^(2+)]["Co" ("CN")_6]^(3-))`

= `"E"_"cell"^circ + 0. 0.59/1 log_10  (["Co"^(3+)]["Co" ("CN")_6]^(4-)["CN"^-]^6)/(["Co"^(2+)]["Co" ("CN")_6]^(3-)["CN"^-]^6)`

= `"E"_"cell"^circ + 0.059/1 log_10  (["Co"^(3+)]["CN"^-]^6)/(["Co"("CN")_6]^(3-)) xx  /((["Co"^(2+)]["CN"]^6)/(["Co" ("CN")_6]^(-4)))`

= `"E"_"cell"^circ + 0.059/1 log_10  ("K"_("f"_1))/("K"_("f"_2)) "K"_("f"_1) = [[("Co" ("CN")_6]^(4-))/(["Co"^(2+)]["CN"^-]^6)`  `["K"_("f"_1) = ("Co"("CN")_6)^(3-)/(["Co"^(3+)][("CN"^-]^6))]`

`0 = 0.83 + 1.82 + 0.059/1 log_10  10^19/"K"_("f"_2)`

`=> "K"_("f"_2)/10^19 = 8.23 xx 10^44`

`=> "K"_("f"_2) = 8.23 xx 10^63`