For a cell involving one electron E^0_{cell} = 0.59V at 298K, the equilibrium constant for the cell reaction is:

[Given that \frac{2.303RT}{F} = 0.059V at T = 298K]

  • Option 1)

    1.0 \times 10^2

  • Option 2)

    1.0\times 10^5

  • Option 3)

    1.0\times 10^{10}

  • Option 4)

    1.0\times 10^{30}

Answers (1)

Option 3 1.0\times 10^{10}

We know,

    E^0_{cell} = \frac{2.303RT}{nF}\log K_{eq}

So,

    \log K_{eq}= \frac{E^0_{cell}}{\frac{2.303RT}{nF}}

    \log K_{eq}= \frac{0.59V}{\frac{0.059}{1}V} \;\;\;\;\;\; [\textup{and}\;n =1 ]

\log K_{eq}= \frac{0.59V}{{0.059}V}

\log K_{eq}=10

\log K_{eq}=10\log_{10}10

\log K_{eq}=\log_{10}10^{10}

K_{eq}=1\times 10^{10}

 


Option 1)

1.0 \times 10^2

Option 2)

1.0\times 10^5

Option 3)

1.0\times 10^{10}

Option 4)

1.0\times 10^{30}

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