Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • Answer please! - Electrochemistry - NEET

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)

best_answer

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}

Posted by

Plabita

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET 2026 LIVE: NTA application correction window closes today; admit card, exam dates, pattern
anam-khan

NEET 2026 registration deadline extended to March 11 for MBBS, BDS; exam on May 3
anam-khan

NEET 2026 Registration LIVE: NTA extends MBBS, BDS application deadline to March 11; how to apply

Latest Question