How much chlorine will be liberated on passing one ampere current

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 Multiple Choice QuestionsMultiple Choice Questions

301.

EMF or hydrogen electrode in terms of pH at 1 atm pressure is

  • EH2 = RTFpH

  • EH2 = RTF1pH

  • EH2 = 2.303 RTFpH

  • EH2 = -0.0591 pH


302.

Assuming that constant current is delivered. How many kW-h of electricity can be produced by the reaction of 1.0 mole Zn with Cu ion in a Daniel cell in which all the concentration remains 1.00 M.

(Given EGiven E°Zn/Zn+=0.76=0.76V),

  • 0.069 kW-h

  • 0.074 kW-h

  • 0.080 kW-h

  • 0.059 kW-h


303.

What amount of Cl2 gas liberated at anode, if 1 ampere current is passed for 30 minute from NaCl solution ?

  • 0.66 mol

  • 0.33 mol

  • 0.66 g

  • 0.33 g


304.

For the given,Sn Sn2+ (0.1 M)  +  2e(Given ESn/Sn2+0= 0.136 V)The electrode potential of the cell at 25° C is:

  • 0.165 V

  • 0750 V

  • 0.421 V

  • 0.012 V


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305.

A 35% solution LiCl was electrolysed by using a 2.5A current for 0.8h.Assuming thecurrent efficiency of 90%. The mass of LiOHproduced at the end of electrolysis.
[Atomic mass of Li=7] is:

  • 1.61 g

  • 2.71 g

  • 4.02 g

  • 3.70 g


306.

Which of the following formula is applicable for weak electrolyte:

  • α=λmλm

  • Ka=2(1-α)

  • Ka=m2λm(λm-λm)

  • All of these


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307.

How much chlorine will be liberated on passing one ampere current for 30 min. through NaCl solution ?

  • 0.66 mole

  • 0.33 mole

  • 0.66 g

  • 0.33 g


C.

0.66 g

According to Faraday's law

wE = it96500or w35.5 = 1 × 30 ×6096500                = 0.622 g

Therefore, 0.662 gm of chlorine is liberated.


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308.

Three ampere currernt was passed through an aqueous solution of an unknown salt of Pd for 1 h 2.977 g of Pdn+ was deposited at cathode.The value of n is:

(given at wf of Pd= 106.4)

  • 6

  • 5

  • 4

  • 3


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309.

The metal which can displace all other metals from their salt solution among the following is:

  • Al

  • Zn

  • Cu

  • Fe


310.

The standard reduction potentials of half reactions are given below:

F2 (g) + 2e-  2F- (aq) ; E0 = + 2.85 V

Cl2 (g) + 2e- 2Cl- (aq) ; E0 = + 1.36 V

Br2 (g) + 2e- 2Br- (aq) ; E0 = + 1.06 V

I2 (g) + 2e-  2I- (aq) ; E= + 0.53 V

The strongest oxidising and reducing agents respectively are:

  • chlorine and iodine

  • fluorine and iodine ion

  • chloride and bromine

  • bromine and iodine


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