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Electrochemistry question

2012 · Q67
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Electrochemistry question

2012 · Q67

NEETChemistryElectrochemistryMCQ+4 / −1
The Gibb's energy for the decomposition of Al2O3 at 500oC is as follows
23{2 \over 3}32​ Al2O3 →\to→ 43{4 \over 3}34​ Al + O2
Δ\DeltaΔrG = +960 kJ mol−-−1
The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at 500oC is at least
  1. A
    4.5 V
  2. B
    3.0 V
  3. C
    2.5 V
  4. D
    5.0 V
View written solutionFree

Correct answer: C

We know,

Δ\Delta ΔGo = – nFEo

23{2 \over 3}32​ Al2O3 →\to→ 43{4 \over 3}34​ Al + O2

Total number of Al atoms in Al2O3

= 23×2=43{2 \over 3} \times 2 = {4 \over 3}32​×2=34​

Al3+ + 3e– →\to→ Al

As 3e– change occur for each Al atom

∴\therefore∴ n = 43×3=4{4 \over 3} \times 3 = 434​×3=4

Eo = - ΔG∘nF{{\Delta G^\circ } \over {nF}}nFΔG∘​

= - 960×10004×96500{{960 \times 1000} \over {4 \times 96500}}4×96500960×1000​

= - 2.5 V

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