Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electrochemistry question

2005 · Shift 0 · Q34
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Electrochemistry
  5. /2005 · Shift 0 · Q34

Electrochemistry question

2005 · Shift 0 · Q34

JEE MainChemistryElectrochemistryMCQ+4 / −1
Aluminium oxide may be electrolysed at 1000oC to furnish aluminium metal (Atomic mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is Al3+Al^{3+}Al3+ + 3e−e^-e− →\to→ Alo To prepare 5.12 kg of aluminium metal by this method would require
  1. A
    5.49 ×\times× 107 C of electricity
  2. B
    1.83 ×\times× 107 C of electricity
  3. C
    5.49 ×\times× 104 C of electricity
  4. D
    5.49 ×\times× 101 C of electricity
View written solutionFree

Correct answer: A

  1. Given data

    • Mass of aluminium to be prepared: 5.12 kg=5120 g5.12\text{ kg} = 5120\text{ g}5.12 kg=5120 g
    • Atomic mass of Al = 27 g mol−127\text{ g mol}^{-1}27 g mol−1
    • Cathode reaction: Al3++3e−→Al\mathrm{Al^{3+} + 3e^- \rightarrow Al}Al3++3e−→Al
    • 111 Faraday =96500 C= 96500\text{ C}=96500 C

  2. Moles of aluminium required n(Al)=512027≈189.63 moln(\mathrm{Al}) = \frac{5120}{27} \approx 189.63\text{ mol}n(Al)=275120​≈189.63 mol

  3. Moles of electrons required From the reaction, 111 mol of Al requires 333 mol of electrons.

    So, n(e−)=3×189.63≈568.89 moln(e^-) = 3 \times 189.63 \approx 568.89\text{ mol}n(e−)=3×189.63≈568.89 mol

  4. Total charge required Since 111 mol of electrons carries 111 Faraday =96500 C= 96500\text{ C}=96500 C, Q=568.89×96500Q = 568.89 \times 96500Q=568.89×96500 Q≈5.49×107 CQ \approx 5.49 \times 10^7\text{ C}Q≈5.49×107 C

  5. Match with options 5.49×107 C\boxed{5.49 \times 10^7\text{ C}}5.49×107 C​ This corresponds to Option A.

  6. Comparison with stored correct answer Stored correct answer: A

    My derived answer: A

    Hence, they agree.

PreviousNext

More from Electrochemistry

  • The highest electrical conductivity of the following aqueous solutions is of :2005 · MCQ
  • Calculate ∧HOAc∞​ Using appropriate molar conductances of the electrolytes listed above at infinite dilution in H2O at 25oC Includes table2005 · MCQ
  • In a hydrogen – oxygen fuel cell, combustion of hydrogen occurs to :2004 · MCQ
  • Consider the following Eo values EFe3+/Fe2+o​= 0.77 V; ESn2+/Sno​= -0.14 V Under standard conditions the potential for the reaction Sn(s) + 2Fe3+(aq) → 2Fe2+(aq) + Sn2+(aq) is :2004 · MCQ
  • 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)2004 · MCQ
  • The limiting molar conductivities Λ° for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol-1 respectively. The Λ° for NaBr is2004 · MCQ
  • In a cell that utilises the reaction Zn(s) + 2H+ (aq) → Zn2+(aq) + H2​(g) addition of H2​SO4​ to cathode compartment, will2004 · MCQ
  • The EM3+/M2+o​ values for Cr, Mn, Fe and Co are – 0.41, +1.57, + 0.77 and +1.97 V respectively. For which one of these metals the change in oxidation state form +2 to +3 is easiest?2004 · MCQ