The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at $25^{o} C$ . The equilibrium constant of the reaction is $(F = 96, 500$ Cmol $^{- 1}, R = 8.314 J K^{- 1}$ mol $^{- 1})$ :

1.0 \times 10^{1}

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1.0 \times 10^{5}

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1.0 \times 10^{10}

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1.0 \times 10^{30}

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Solution

The correct option is C $1.0 \times 10^{10}$
Using the relation:

$n F E = R T l n (K_{e q})$

$1 \times 96500 \times 0.591 = 8.314 \times 298 \times 2.303 \times log K_{e q}$

$log K_{e q} = 9.995$

$K_{e q} = 10^{9.995} \approx 10^{10}$

$∴$ C option is correct

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Similar questions

The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at $25^{0}$ C. The equilibrium constant of the reaction is (F = 96500Cmol $^{- 1}$ , R = 8.314JK $^{- 1}$ mol $^{- 1}$ ) :

Q. The standard emf of a cell, involving one electron change is found to be

0.591 V

25

. The equilibrium constant of the reaction is

(F = 96500 C {m o l}^{- 1}, R = 8.314 K^{- 1} {m o l}^{- 1})

Q. For a cell involving one electron,

E_{c e l l}^{0}

= 0.59 V at 298 K, the equilibrium constant for the cell reaction is:
[Given that

\frac{2.303 R T}{F}

= 0.059 V at T = 298 K]

Q. The standard emf of a cell involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is

[Given:

F = 96500 C m o l^{- 1}

]

Q. The standard e.m.f. of a cell involving one electron change is found to 0.591 V at 25

^{o}

C. The equilibrium constant of the reaction is:

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The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at 25oC. The equilibrium constant of the reaction is (F=96,500 Cmol−1,R=8.314JK−1 mol−1):

The correct option is C 1.0×1010Using the relation:nFE=RT ln(Keq)1×96500×0.591=8.314×298×2.303×logKeq logKeq=9.995Keq=109.995≈1010∴ C option is correct

The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at $25^{o} C$ . The equilibrium constant of the reaction is $(F = 96, 500$ Cmol $^{- 1}, R = 8.314 J K^{- 1}$ mol $^{- 1})$ :

The correct option is C $1.0 \times 10^{10}$
Using the relation:

$n F E = R T l n (K_{e q})$

$1 \times 96500 \times 0.591 = 8.314 \times 298 \times 2.303 \times log K_{e q}$

$log K_{e q} = 9.995$

$K_{e q} = 10^{9.995} \approx 10^{10}$

$∴$ C option is correct