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Electrochemistry

Question

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Calculate emf of the following cell at 298 K
Mg(s) | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu(s)
(Given : E°_Cu2/Cu = + 0.34 V; E° _Mg²⁺/Mg = – 2.37 V)

Solution

Mg (s) \to {Mg}^{2 +} + 2 e^{-}

{Cu}^{2 +} + 2 e^{-} \to Cu (s) The overall net reaction Mg (s) + {Cu}^{2 +} \to {Mg}^{2 +} + Cu (s) E_{cell} = E_{cell}^{0} - \frac{2.303}{nF} RT \log \frac{[{Mg}^{2 +}]}{[{Cu}^{2 +}]} E_{cell} = E_{\frac{{Cu}^{2 +}}{Cu}}^{0} - E_{\frac{{mg}^{2 +}}{mg}}^{0} = + 0.34 V - (- 2.37 V) = + 2.71 V E_{cell} = 2.71 - \frac{0.0591}{2} \times \log 10 E_{cell} = 2.71 V - 0.0295 E_{cell} = 2.6805 V

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