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Electrochemistry question

2020 · 5 Sep · Shift 1 · Q15
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  5. /2020 · 5 Sep · Shift 1 · Q15

Electrochemistry question

2020 · 5 Sep · Shift 1 · Q15

JEE MainChemistryElectrochemistryNumerical+4 / −1
An oxidation-reduction reaction in which 3 electrons are transferred has a Δ\DeltaΔ Gº of 17.37 kJ mol–1 at 25 oC. The value of Eo cell (in V) is ‾\underline{\hspace{2cm}}​ × 10–2. (1 F = 96,500 C mol–1)
Numerical answer
View written solutionFree

Correct answer: -6

  1. Use the relation between standard Gibbs free energy and standard cell potential:

ΔG∘=−nFEcell∘\Delta G^\circ = -n F E^\circ_{\text{cell}}ΔG∘=−nFEcell∘​

Given:

  • ΔG∘=17.37 kJ mol−1=17.37×103 J mol−1\Delta G^\circ = 17.37\,\text{kJ mol}^{-1} = 17.37 \times 10^3\,\text{J mol}^{-1}ΔG∘=17.37kJ mol−1=17.37×103J mol−1
  • n=3n=3n=3
  • F=96500 C mol−1F=96500\,\text{C mol}^{-1}F=96500C mol−1
  1. Rearrange to find Ecell∘E^\circ_{\text{cell}}Ecell∘​:

Ecell∘=−ΔG∘nFE^\circ_{\text{cell}} = -\frac{\Delta G^\circ}{nF}Ecell∘​=−nFΔG∘​

Substitute values:

Ecell∘=−17.37×1033×96500E^\circ_{\text{cell}} = -\frac{17.37 \times 10^3}{3 \times 96500}Ecell∘​=−3×9650017.37×103​

  1. Calculate denominator:

3×96500=2895003 \times 96500 = 2895003×96500=289500

So,

Ecell∘=−17370289500=−0.06 VE^\circ_{\text{cell}} = -\frac{17370}{289500} = -0.06\,\text{V}Ecell∘​=−28950017370​=−0.06V

  1. Express in the required form:

−0.06 V=−6×10−2 V-0.06\,\text{V} = -6 \times 10^{-2}\,\text{V}−0.06V=−6×10−2V

Hence, the blank is filled by:

−6\boxed{-6}−6​

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