+3 votes
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in JEE by (179 points)
a)0.29 L

b)0.28 L

c)0.20 L

d)0.3 L

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2 Answers

+2 votes
by (64.3k points)

Solution: 

  • Electrode equations:

    • (-) cathode 2H+ + 2 e- ==> H2 or 4H+ + 4e- ==> 2H2 for molar electron comparison

    • (+) anode 4OH- -2e- ==> 2H2O(l) + O2 

  • Quantity of electricity in Coulombs = current in A x time in seconds

  • Q = I x t = 1 x 20 x 60 = 1200 C, now 1 mole of electrons = 96500 C

  • so moles of electrons = 1200 / 96500 = 0.012435 moles

  • it takes 4 moles of electrons to form 1 mole of oxygen gas

  • therefore moles of oxygen formed = 0.012435 / 4 = 0.0031

  • 1 mole of gas = 24000 cm3, therefore volume of gas = 0.0031 x 24000 = 74.4 cm3 of O2

  • volume of gas = 0.0744 L of O2

by (179 points)
But there is no such option????
by (64.3k points)
@Team B.I  As per your question, that is the answer.
You can check your question again
by (179 points)
The question is correct
by (64.3k points)
yes, In question it asked total volume of gases and i have solved for volume of oxygen
You can get the volume of Hydrogen in the same manner and hence
total volume of gases will be volume of oxygen + volume of hydrogen
by (179 points)
But I am not understanding what have you done? Please can you explain with some equations?
by (64.3k points)
Read the concept i have provided in second answer.
Hope this helps you
+1 vote
by (64.3k points)

Concept to solve the above question:

  • As we know the amount of product formed is proportional to time and current.
  • This is expressed by the following formula ...
    • Quantity of charge transferred (Coulombs, C) = current flow (amperes, A) x time (seconds, s)
    • Q = I x t
    • The amount of product formed is proportional to Q, the charge transferred to oxidise or reduce ions in the electrolyte.
    • So you can logically deduce the following sort of proportionality from the Q = It equation ..
      • Doubling the current with double the rate of production etc., halving the current halves the rate ... etc.
      • Doubling the time will double the amount of electrode products formed etc., halving the time will halve the amount of product formed ... etc.
      • Substituting numbers into the equation enables you to predict how charge flows, then by a subsequent calculation, you can predict by calculation how much product will be formed e.g.
        • e.g. if you run a current of 1.50 A for 2 minutes, how charge has flowed?
        • Q = I x t = 1.50 x (2 x 60) = 180 C
        • The 180 Coulombs can be converted into moles of electrons, and then, adjusting for the charge on the ion, converted into moles of ion discharged at the electrode, and finally moles of product.
        • The following examples show you how to do all these calculations in a logical manner.
        • You may also need to rearrange the equation to determine current needed or time taken.
        • Q = I x t,   I = Q / t  and  t = Q / I   (do we need a triangle guys?)
        • How long would it take to pass 5000 Coulombs, with a current of 4A?
        • t = Q / I = 5000 / 4 = 1250 s (or 20.83 minutes, 20 mins & 50 seconds)
        • What current do you need to pass 96500 Coulombs in 2 hours?
        • I = Q / t  = 96500 / (20 x 60 x 60) = 1.34A
  • 1 Faraday (F) = 96 500 Coulombs (C) = 1 mole of electrons.

    • This can be expressed as the Faraday Constant = 96500 Cmol-1

    • A current of 1A = 1C/s

    • 1 mole of any gas = 24 dm3 or 24000cm3 at 25oC/1 atmosphere.

  • Quantity of electricity in coulombs current in amps x time in seconds

    • Q (C) = I (A) x t (s)

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