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Electrochemistry

Question

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Calculate the equilibrium constant for the reaction:
Cd²⁺(aq) + Zn(s) → Zn²⁺(aq) + Cd(s)
if E°_Cd²⁺/Cd = – 0.403 V.
E°_zn²⁺/zn = – 0.763 V

Solution

{E^{0}}_{cell} = {E^{0}}_{({Cd}^{2 +} / Cd)} - {E^{0}}_{({Zn}^{2 +} / Zn)} = - 0.403 - (0.763) = 0.36 V {E^{0}}_{cell} = \frac{0.059}{n} \log K \log K = \frac{n + {E^{0}}_{cell}}{0.059} \log K = \frac{2 + 0.36}{. 059} = 12.2 or K = 1.585 \times 10^{12} .

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