Given:(i) Cu^2+ + 2e^- → Cu , E^= 0.337 V (ii) Cu^2+ + e^- → Cu^+ , E^= 0.153 V Electrode potential, E^ for the reaction, Cu^+ + e^- → Cu , will be: - Chemistry MCQ Chemistry | ExamTest.live

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Given:(i) Cu^{2 +} + 2 e^{-} \to Cu, E^{\circ} = 0.337 V (ii) Cu^{2 +} + e^{-} \to Cu^{+}, E^{\circ} = 0.153 V Electrode potential, E^{\circ} for the reaction, Cu^{+} + e^{-} \to Cu, will be:

A $0.90 V$
B $0.30 V$
C $0.38 V$
D $0.52 V$

Solution & Step-by-step Explanation

Potentials are not additive, but Gibbs free energy is. Δ G^{\circ} = - n F E^{\circ} .Let reaction (iii) be Cu^{+} + e^{-} \to Cu .Notice: Eq(i) = Eq(ii) + Eq(iii) Therefore: Δ G_{1}^{\circ} = Δ G_{2}^{\circ} + Δ G_{3}^{\circ} $ - n_{1} F E_{1}^{\circ} = - n_{2} F E_{2}^{\circ} - n_{3} F E_{3}^{\circ} $ 2 (0.337) = 1 (0.153) + 1 (E_{3}^{\circ}) 0.674 = 0.153 + E_{3}^{\circ} E_{3}^{\circ} = 0.674 - 0.153 = 0.521 V

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Given:(i) Cu^{2 +} + 2 e^{-} \to Cu, E^{\circ} = 0.337 V (ii) Cu^{2 +} + e^{-} \to Cu^{+}, E^{\circ} = 0.153 V Electrode potential, E^{\circ} for the reaction, Cu^{+} + e^{-} \to Cu, will be:

0.90 V

0.30 V

0.38 V

0.52 V

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Electrochemistry Electrode Potential Gibbs Free Energy

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Given:(i) $Cu^{2 +} + 2 e^{-} \to Cu$ , $E^{\circ} = 0.337 V$ (ii) $Cu^{2 +} + e^{-} \to Cu^{+}$ , $E^{\circ} = 0.153 V$ Electrode potential, $E^{\circ}$ for the reaction, $Cu^{+} + e^{-} \to Cu$ , will be:

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Given:(i) Cu2++2e−→Cu, E∘=0.337V (ii) Cu2++e−→Cu+, E∘=0.153V Electrode potential, E∘ for the reaction, Cu++e−→Cu, will be:

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Given:(i) $Cu^{2 +} + 2 e^{-} \to Cu$ , $E^{\circ} = 0.337 V$ (ii) $Cu^{2 +} + e^{-} \to Cu^{+}$ , $E^{\circ} = 0.153 V$ Electrode potential, $E^{\circ}$ for the reaction, $Cu^{+} + e^{-} \to Cu$ , will be: