What amount of Cl2 gas liberated at anode, if 1 ampere current is

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 Multiple Choice QuestionsMultiple Choice Questions

631.

Electrolysis of fused NaCl will give

  • Na

  • NaOH

  • NaClO

  • NaClO3


632.

How long it will take to deposit 1.0 g of chromium when a current of 1.25 A flows through a solution of chromium (III) sulphate ? (Molar mass of Cr = 52)

  • 1.24 min

  • 1.24 h

  • 1.24 s

  • None of these


633.

The compounds formed at anode in the electrolysis of an aqueous solution of potassium acetate, are

  • C2H6 and CO2

  • C2H4 and CO2

  • CH4 and H2

  • CH4 and CO2


634.

What is the electrode potential (in V) of the following electrode at 25°C ?

Ni2+ (0.1 M) | Ni(s)

(Standard reaction potential of Ni2+ | Ni is -0.25V, 2.303 RTF = 0.06)

  • -0.28V

  • -0.34V

  • -0.82V

  • -0.22V


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635.

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


636.

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


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637.

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


C.

0.66 g

Amount of Cl2 gas liberated at anode, if 1 ampere current is passed for 30 minutes from NaCl solution is 0.66 gm.

2Cl-  Cl21 mole + 2e- (At anode)2 x 96500 coulomb

Q = it = 1 x 30 x 60

   = 1800 coulomb

The amount of chlorine liberated by passing 1800 coulomb of electric charge

1 × 1800 × 712 × 96500 = 0.66 gm


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638.

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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639.

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


640.

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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