The equilibrium constant of the reaction: Cu ( s ) + 2 Ag ( a q ) + → Cu ( a q ) 2 + + 2 Ag ( s ) E o = 0.46 V at 298 K is:NEET 2007

As E cell 0 = n 0.0591 lo g K c ∴ 0.46 = 2 0.0591 lo g K c ∴ lo g K c = 0.0591 2 × 0.46 = 15.57 or K c = Antilog 15.57 = 3.7 × 1 0 15

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Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsHard2 minPYQ_2007

ChemistryHardsingle choice

The equilibrium constant of the reaction: $Cu_{(s)} + 2 Ag_{(a q)}^{+} \to Cu_{(a q)}^{2 +} + 2 Ag_{(s)}$ $E^{o} = 0.46 V$ at $298 K$ is:NEET 2007

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Answer:

Solution:

As $E_{cell}^{0} = \frac{0.0591}{n} lo g K_{c}$ $∴ 0.46 = \frac{0.0591}{2} lo g K_{c} ∴ lo g K_{c} = \frac{2 \times 0.46}{0.0591} = 15.57 or K_{c} = Antilog 15.57 = 3.7 \times 1 0^{15}$

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

⏱ 2mℹ️ Source: PYQ_2007

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