Given E Cu 2 + / Cu ∘ = 0.34 V , E Cu 2 + / Cu ∘ = 0.15 V Standard electrode potential for the half cell Cu + / Cu is

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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsHard2 minPYQ_2012

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Given $E_{Cu^{2 +} / Cu}^{\circ} = 0.34 V, E_{Cu^{2 +} / Cu}^{\circ} = 0.15 V$ Standard electrode potential for the half cell $Cu^{+} / Cu$ is

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$Cu^{++} + e^{-} ⟶ Cu^{+}; E_{1}^{0} = 0.15 V; Δ G_{1}^{0} = - n_{1} E_{1}^{0} F Cu^{2 +} + 2 e ⟶ Cu; E_{2}^{0} = 0.34 V; Δ G_{2}^{0} = - n_{2} E_{2}^{0} F$ On subracting eq.(i) from eq. (ii) we get $Cu^{+} + e^{-} ⟶ Cu; Δ G^{0} = Δ G_{2}^{0} - Δ G_{1}^{0} - n E^{0} F = - (n_{2} E^{0} F - n_{1} E_{1}^{0} F) E^{0} = \frac{n _{2} E _{2}^{0} F - n _{1} E _{1}^{0} F}{n F} = \frac{2 \times 0.34 - 0.15}{1} = 0.53 V$

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

⏱ 2mℹ️ Source: PYQ_2012

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