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Electrochemistry question

2003 · Shift 0 · Q33
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Electrochemistry question

2003 · Shift 0 · Q33

JEE MainChemistryElectrochemistryMCQ+4 / −1
For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295 V at 25oC. The equilibrium constant of the reaction at 25oC will be
  1. A
    29.5 ×\times× 10-2
  2. B
    10
  3. C
    1 ×\times× 1010
  4. D
    1 ×\times× 10-10
View written solutionFree

Correct answer: C

  1. Use the relation between standard emf and equilibrium constant

For a cell reaction,

ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circΔG∘=−nFE∘

and also,

ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln KΔG∘=−RTlnK

Equating both:

RTln⁡K=nFE∘RT\ln K = nFE^\circRTlnK=nFE∘

So,

ln⁡K=nFE∘RT\ln K = \frac{nFE^\circ}{RT}lnK=RTnFE∘​

At 25∘25^\circ25∘C, this is commonly written as:

E∘=0.0591nlog⁡KE^\circ = \frac{0.0591}{n}\log KE∘=n0.0591​logK

Hence,

log⁡K=nE∘0.0591\log K = \frac{nE^\circ}{0.0591}logK=0.0591nE∘​
  1. Substitute the given values

Given:

  • n=2n=2n=2
  • E∘=0.295 VE^\circ = 0.295\,\text{V}E∘=0.295V

So,

log⁡K=2×0.2950.0591\log K = \frac{2 \times 0.295}{0.0591}logK=0.05912×0.295​ log⁡K=0.5900.0591≈9.98≈10\log K = \frac{0.590}{0.0591} \approx 9.98 \approx 10logK=0.05910.590​≈9.98≈10
  1. Find KKK
K=1010K = 10^{10}K=1010
  1. Match with the options

This corresponds to:

Option C: 1×10101 \times 10^{10}1×1010

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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