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Electrochemistry

Question

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Calculate the standard free energy change for the following reaction at 25°C.
Au(s) + Ca²⁺(aq.1M) → Au³⁺(aq. 1M) + Ca(s)
E°_Au²⁺/Au = – 0.403 V.
E°_Au²⁺/Cu = – 0.763 V.
Predict whether the reaction will be spontaneous or not at 25°C. Which of the above two half cells will act as an oxidizing agent and which one will be a reducing agent?

Solution

E ° = E °_{{Ca}^{2 +} / Ca} - E °_{{Au}^{3 +} / Au} = - 287 - (1.50) = - 2.87 - 1.50 = 4.37 V ∆ rG ° = nFE °

= 6 \times 96500 \times (- 4.37) = 2530.23 kg

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