For the disproportionation reaction 2 Cu + aq ⇌ Cu s + Cu 2 + aq at 298 K , ln   K (where K is the equilibrium constant) is _______ × 10 - 1 Given : E ° Cu 2 + / Cu + = 0 . 16 V   E ° Cu + / Cu = 0 . 52 V RT F

Question

For the disproportionation reaction 2 Cu + aq &#8652; Cu s + Cu 2 + aq at 298 K , ln &#160; K (where K is the equilibrium constant) is _______ &#215; 10 - 1 Given : E &#176; Cu 2 + / Cu + = 0 . 16 V &#160; E &#176; Cu + / Cu = 0 . 52 V RT F = 0 . 025

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E cell o = E Cu + / Cu &#176; - E Cu 2 + / Cu + &#176; = 0 . 52 - 0 . 16 = 0 . 36 &#160; V E cell &#176; = RT nF lnK eq 0 . 36 = 0 . 025 1 lnk In &#160; k = 14 . 4 &#160; &#160; &#160; &#160; &#160; &#160; = 144 &#215; 10 - 1

For the disproportionation reaction 2 Cu + aq ⇌ Cu s + Cu 2 + aq at 298 K , ln   K (where K is the equilibrium constant) is _______ × 10 - 1 Given : E ° Cu 2 + / Cu + = 0 . 16 V   E ° Cu + / Cu = 0 . 52 V RT F = 0 . 025

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For the disproportionation reaction 2 Cu + aq &#8652; Cu s + Cu 2 + aq at 298 K , ln &#160; K (where K is the equilibrium constant) is _______ &#215; 10 - 1 Given : E &#176; Cu 2 + / Cu + = 0 . 16 V &#160; E &#176; Cu + / Cu = 0 . 52 V RT F = 0 . 025

Correct answer

Step-by-step solution

More from Electrochemistry

For the disproportionation reaction 2 Cu + aq ⇌ Cu s + Cu 2 + aq at 298 K , ln K (where K is the equilibrium constant) is _______ × 10 - 1 Given : E ° Cu 2 + / Cu + = 0 . 16 V E ° Cu + / Cu = 0 . 52 V RT F = 0 . 025