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Electrochemistry question

2004 · Shift 0 · Q37
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Electrochemistry question

2004 · Shift 0 · Q37

JEE MainChemistryElectrochemistryMCQ+4 / −1
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 C mol-1: R = 8.314 JK-1 mol-1)
  1. A
    1.0 ×\times× 101
  2. B
    1.0 ×\times× 1030
  3. C
    1.0 ×\times× 1010
  4. D
    1.0 ×\times× 105
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 = -n F E^\circΔG∘=−nFE∘

and also,

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

Equating both:

−nFE∘=−RTln⁡K-n F E^\circ = -RT \ln K−nFE∘=−RTlnK ⇒ln⁡K=nFE∘RT\Rightarrow \ln K = \frac{nFE^\circ}{RT}⇒lnK=RTnFE∘​
  1. Substitute the given values

Given:

  • n=1n=1n=1
  • E∘=0.591 VE^\circ = 0.591\,\text{V}E∘=0.591V
  • F=96500 C mol−1F = 96500\,\text{C mol}^{-1}F=96500C mol−1
  • R=8.314 J K−1mol−1R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}R=8.314J K−1mol−1
  • T=25∘C=298 KT = 25^\circ \text{C} = 298\,\text{K}T=25∘C=298K

So,

ln⁡K=(1)(96500)(0.591)(8.314)(298)\ln K = \frac{(1)(96500)(0.591)}{(8.314)(298)}lnK=(8.314)(298)(1)(96500)(0.591)​

First calculate numerator:

96500×0.591=57031.596500 \times 0.591 = 57031.596500×0.591=57031.5

Now denominator:

8.314×298=2477.5728.314 \times 298 = 2477.5728.314×298=2477.572

Thus,

ln⁡K=57031.52477.572≈23.0\ln K = \frac{57031.5}{2477.572} \approx 23.0lnK=2477.57257031.5​≈23.0
  1. Convert from natural log to common exponential form
K=e23.0K = e^{23.0}K=e23.0

Using ln⁡10=2.303\ln 10 = 2.303ln10=2.303,

log⁡10K=23.02.303≈10.0\log_{10} K = \frac{23.0}{2.303} \approx 10.0log10​K=2.30323.0​≈10.0

Hence,

K=1010K = 10^{10}K=1010
  1. Match with the options

The correct option is:

C: 1.0×1010\boxed{\text{C: } 1.0 \times 10^{10}}C: 1.0×1010​
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