Define fuel cell and write its two advantages. from Chemistry El

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 Multiple Choice QuestionsShort Answer Type

331.

Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes.
(Given : Molar mass of Ag = 108 g mol–1 1F = 96500 C mol–1)

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332.

Define fuel cell.

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333.

For the reaction

2AgCl (s) + H2 (g)  (1 atm)   2 Ag (s) + 2H+ (0.1 M) + 2Cl- (0.1 M),Go = - 43600 J at 25o C

Calculate the e.m.f. of the cell.
[log 10–n = – n]


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334.

Define fuel cell and write its two advantages.


These are voltaic cells in which the reactants are continuously supplied to the electrodes. There are designed to convert the energy from the combustion of a fuel like a hydrogen, methane, methanol, etc directly into electrical energy.
Advantage : -
(1) A fuel cell works with an efficiency of 60 to 70 %
(2) They are pollution free


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 Multiple Choice QuestionsMultiple Choice Questions

335.

Which among the following metals is employed to provide cathodic protection to iron?

  • Zinc

  • Nickel

  • Tin

  • Lead


336.

Galvanization is applying a coating of:

  • Cr

  • Cu

  • Zn

  • Zn

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337.

Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is: (at. mass of Cu = 63.5 amu)

  • 0 g

  • 63.5 g

  • 2 g

  • 2 g

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338.

The equivalent conductance of NaCl at concentration C at infinite dilution are λC and λ respectively. The correct relationship between λC and λ  is given as (where the constant B is positive)

  • λC = λ +(B)C

  • λC = λ -(B)C

  • λC = λ -(B)square root of straight C

  • λC = λ -(B)square root of straight C

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339.

Given below are the half-cell reactions
Mn2+ + 2e- → Mn; Eo = - 1.18 eV
2(Mn3+ + e- →Mn2+); Eo = +1.51 eV
The Eo for 3Mn2+ → Mn + 2Mn3+ will be

  • -2.69 V; the reaction will not occur

  • -2.69 V; the reaction will occure

  • -0.33 V; the reaction will not occur

  • -0.33 V; the reaction will not occur

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340.

Given, 

straight E subscript Cr to the power of 3 plus end exponent divided by Cr end subscript superscript straight o space equals space minus space 0.74 space straight V semicolon
straight E subscript MnO subscript 4 superscript minus divided by Mn to the power of 2 plus end exponent end subscript superscript straight o space equals space 1.51 space straight V
straight E subscript Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript divided by Cr to the power of 3 plus end exponent end subscript superscript straight o space equals space 1.33 space straight V semicolon
straight E subscript Cl divided by Cl to the power of negative 1 end exponent end subscript superscript straight o space equals space 1.36 space straight V
Based on the data given above, strongest oxidising agent will be

  • Cl

  • Cr3+

  • Mn2+

  • Mn2+

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