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Calculate the equilibrium constant for the reactant at 298 K Cu (s) + 2 Ag^(+) (aq) to Cu^(2+) (aq) + 2 Ag (s) Given that E_((Ag^(+) |Ag))^(Theta) = 0.80 Vand E_((Cu^(2+) |Cu))^(Theta) = 0.34 V . |
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Answer» Solution :`E_(cell)^(Theta)` is RELATED to equilibrum constant `K_(E)` as : 298 K as: `E_(cell)^(Theta) = (0.059)/(n) log K_(c) ` or `log K_(c) = (n E_(cell)^(Theta))/(0.059)` `E_(cell)^(Theta) = E_((Ag^(+) | Ag))^(Theta) - E_((Cu^(2+) |Cu))^(Theta)` `= 0.80 - 0.34 = 0.46` V or log `K_(c) = (2 xx 0.46)/(0.059) = 15.59` `THEREFORE "" K_(c) = 3.92 xx 10^(15)` . |
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