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Electrochemistry

Question

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Calculate the equilibrium constant for the reaction at 298 K
$Cu (s) + {Cl}_{2} (g) ⇌ {CuCl}_{2} (aq)$
Given: $E °_{{Cu}^{2 +} / Cu} = 0.34 V; E °_{{Cl}^{2 +} / {Cl}^{-}} = 1.36 V, R = 8.314 J K^{- 1} {mol}^{- 1}$

Solution

R = 96500 C {mol}^{- 1} E °_{cell} = E °_{cathode} - E °_{anode} = 1.36 - 0.34 = 1.02 V

Equilibrium constant,

\log K_{C} = E °_{cell} \times \frac{nF}{2.303 RT}

or

\log K_{C} = \frac{1.02 \times 2 \times 96500}{2.303 \times 8.314 \times 298} = 34.57

or

K_{C} = Antilog (34.57) = 3.715 \times 10^{34}

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