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Electrochemistry

Question

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Calculate the equilibrium constant of the reaction
$Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + 2 Ag (s)$
given : E⁰ =0.46 temperture 25⁰ C

Solution

we have given that:

E ° = + 0.46 V

E ° = \frac{RT}{nF} lnK lnK = \frac{nFE °}{RT}

lnk =2 x96500 x0.46/0.059

K = antilog (\frac{nE °}{0.059})

K = antilog (\frac{2 \times 0.46}{0.059})

K = antilog (15.5932) K = 3.92 \times 10^{15} .

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