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Electrochemistry question

2019 · Q83
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Electrochemistry question

2019 · Q83

NEETChemistryElectrochemistryMCQ+4 / −1
For a cell involving one electron EcellΘE_{cell}^\ThetaEcellΘ​ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is :

[Given that 2.303RTF{{2.303RT} \over F}F2.303RT​ = 0.059 V at T = 298 K ]
  1. A
    1.0 ×\times× 1030
  2. B
    1.0 ×\times× 1010
  3. C
    1.0 ×\times× 102
  4. D
    1.0 ×\times× 105
View written solutionFree

Correct answer: B

We know,

Ecell = EcellΘE_{cell}^\Theta EcellΘ​ - 2.303RTnFlog⁡Q{{2.303RT} \over {nF}}\log QnF2.303RT​logQ

⇒\Rightarrow⇒ Ecell = EcellΘE_{cell}^\Theta EcellΘ​ - 0.059nlog⁡Q{{0.059} \over n}\log Qn0.059​logQ

(At equilibrium, Ecell = 0 and Q = Keq)

⇒\Rightarrow⇒ 0 = EcellΘE_{cell}^\Theta EcellΘ​ - 0.0591log⁡Keq{{0.059} \over 1}\log {K_{eq}}10.059​logKeq​

⇒\Rightarrow⇒ log⁡Keq=EcellΘ0.059\log {K_{eq}} = {{E_{cell}^\Theta } \over {0.059}}logKeq​=0.059EcellΘ​​ = 0.590.059{{0.59} \over {0.059}}0.0590.59​ = 10

⇒\Rightarrow⇒ Keq = 1010

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