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Electrochemistry question

2003 · Q101
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Electrochemistry question

2003 · Q101

NEETChemistryElectrochemistryMCQ+4 / −1
On the basis of the information available from the reaction,

4/3Al + O2 →\to→ 2/3Al2O3,   Δ\DeltaΔG = −-− 827 kJ mol−-−1 of O2,

the minimum e.m.f. required to carry out an electrolysis of Al2O3 is
(F = 96500 C mol−-−1)
  1. A
    2.14 V
  2. B
    4.28 V
  3. C
    6.42 V
  4. D
    8.56 V
View written solutionFree

Correct answer: A

4/3Al + O2 →\to→ 2/3Al2O3,   Δ\Delta ΔG = −-− 827 kJ mol−-−1

For 1 mol of Al, n = 3

∴\therefore∴ For 43{4 \over 3}34​ mol of Al, n = 3×43=43 \times {4 \over 3} = 43×34​=4

As Δ\Delta ΔG = - nFEo

⇒\Rightarrow⇒ – 827 × 103 J = – 4 × E° × 96500

⇒\Rightarrow⇒ Eo = 827×1034×96500{{827 \times {{10}^3}} \over {4 \times 96500}}4×96500827×103​ = 2.14 V

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