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प्रश्न
For a cell reaction in involving a two electron change, the Stanford emf of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be
पर्याय
2.95 × 10–2
10
1 × 10–10
2.95 × 10–10
1 × 1010
MCQ
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उत्तर
10 or 1 × 1010
Explanation:
Using the relation,
`E_(cell)^circ = (2.303 RT)/(nF) log K_c`
= `0.0591/n log K_c`
∴ 0.295 V = `0.0591/2 log K_c`
or `log K_c = (2 xx 0.295)/0.0591`
Kc = 10
or Kc = 1 × 1010
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Notes
Students should refer to the answer according to their options.
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