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Electrochemistry

Question

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Calculate the equilibrium constant for the reaction,
Cu(s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag(s)
[E°_cell = 0.46 V]
(T = 298 K, F = 96500 C mol^–1, R = 8.31 J K^–1 mol^–1)

Solution

Using the formula we get:

∆ G ° = - nFE °_{cell} = - 2.303 RT \log K \log K = \frac{nFE °_{cell}}{2.303 \times R \times T} = \frac{2 \times 96500 \times 0.46}{2.303 \times 8.31 \times 298} = 15.5 K = 3.69 \times 10^{15} .

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