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Electrochemistry

Question

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Write the Nernst equation and calculate the emf of the following cell at 298 K : Cu(s) | Cu²⁺ (0.130 M) || Ag⁺ (1.00 x 10^–4 M) | Ag(s)

Solution

E°_Cu²⁺/ cu = + 0.34 V and E°_Ag+/Ag = + 0.80 V.

Cu + 2 {Ag}^{+} \to {Cu}^{2 +} + 2 Ag (s) A p p l y i n g t h e n e r n s t e q u a t i o n E = E ° - \frac{0.059}{2} \log \frac{[{Cu}^{2 +}]}{{[{Ag}^{+}]}^{2}} E ° = E °_{cathode} - E °_{anode} = 0.80 - 0.34 = 0.46 V E = 0.46 - \frac{0.059}{2} \log \frac{0.130}{1.0 \times 10^{- 4}} = 0.46 - \frac{0.059}{2} \times 1.30 \times 10^{7} = 0.46 - 0.21 = 0.25 V .

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