Given the equilibrium constant: K C of the reaction: Cu ( s ) + 2 Ag + ( aq ) → Cu 2 + ( aq ) + 2 Ag ( s ) is 10 × 1 0 15 calculate the E cell 0 of this reaction at 298 K [ 2.303 F RT at 298 K = 0.059 V ]

E cell 0 = nF 2.303 RT lo g K C or E cell 0 = n 0.059 V lo g K C = 2 0.059 V lo g 1 0 16 = 0.4736 V

Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsHard2 minPYQ_2019

ChemistryHardstatement

Given the equilibrium constant: $K_{C}$ of the reaction: $Cu (s) + 2 Ag^{+} (aq) \to Cu^{2 +} (aq) + 2 Ag (s)$ is $10 \times 1 0^{15}$ calculate the $E_{cell}^{0}$ of this reaction at $298 K$ $[2.303 \frac{RT}{F}$ at $298 K = 0.059 V]$

Options:

Answer:

Solution:

$E_{cell}^{0} = \frac{2.303 RT}{nF} lo g K_{C}$ or $E_{cell}^{0} = \frac{0.059 V}{n} lo g K_{C}$ $= \frac{0.059 V}{2} lo g 1 0^{16} = 0.4736 V$

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

⏱ 2mℹ️ Source: PYQ_2019

Doubts & Discussion

Loading discussions...