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Electrochemistry

Question

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Calculate the cell emf and AG for the cell rection at 25°C for the cell
Zn(s) | Zn²⁺ (0.0004 M) || Cd²⁺ (0.2 M) | Cd(s) E° values at 25°C, Zn²⁺/Zn = – 0.763 V, Cd²⁺/ Cd = – 0.403 V F = 96500 C mol^–1, R = 8.314 J K^–1 mol^–1

Solution

Applying nernst equation :

E °_{cell} = E °_{cathode} - E ° anode = - 0.403 - (- 0.763) = 0.36 V

E_{cell} = E °_{cell} + \frac{0.059}{n} \log \frac{[{Cd}^{2 +}]}{[{Zn}^{2 +}]}

or

E_{cell} = 0.36 + \frac{0.059}{n} \log \frac{0.2}{0.004} = 0.36 + \frac{0.059}{2} \log 500 = 0.36 + \frac{0.059}{2} \times 2.6990 E_{cell} = 0.4396 V

Now

∆_{r} G = - nFE

or

∆_{r} G = - 2 \times 96500 \times 0.4396 = - 84842.8 J {mol}^{- 1} = 84.84 kJ {mol}^{- 1}

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