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Water electrolysis - solving for final volume of the system

  1. Jul 10, 2015 #1
    1. The problem statement, all variables and given/known data
    2 moles of water at 20°C undergoes electrolysis at 5A/40V inside of a closed 40mL container for 1 min at standard external/internal pressure. What is the final volume of the system?

    2. Relevant equations

    (Q*M)/(F*z)=m

    Q=c x m x ΔT

    Vi/Ti = Vf/Tf


    3. The attempt at a solution

    Gas Produced
    (5A*60s*4g)/(F*4)=0.00311g H2
    (5A*60s*32g)/(F*4)=0.024875g O2
    0.00311g + 0.024875g = 0.02798g

    Temperature of the system
    Q=cmΔT
    5A*40V*60s=12kJ
    Liquid 36g x 4.18J/(g⋅K)=150.48J/(g⋅K)
    Gas 0.02798g x 2.080J/(g⋅K)=0.0582J/(g⋅K)
    150.48J/(g⋅K) + 0.0582J/(g⋅K)=150.538J/(g⋅K)
    12kJ=150.538J/(g⋅K)/(g⋅K)*ΔT
    12kJ/150.538J/(g⋅K)=(T2-20°C)
    79.7 + 20°C=99.7°C

    Does the rise in temperature cause the total volume of the system to increase? If so, how do I calculate the change in volume? I think I'm supposed to use the volume of the container as initial volume and use Charle's law to calculate the change but I'm not sure in this case.

    Vi/Ti = Vf/Tf
    0.040L/293K=Vf/372.85
    Vf=0.05L
     
  2. jcsd
  3. Jul 11, 2015 #2
    the volume of gaseous products in the container is determined by the quantity of water. initially it is about 3.96 ml and after electolysis is will be slightly greater due to the change of liquid to gaseous products Water at NTP is considered incompressible so pressure can be neglected. Rising temp will expand the water but also the container. Not having info on the container we should neglect this effect too

    .So how much water in converted?
     
    Last edited by a moderator: Jul 11, 2015
  4. Jul 11, 2015 #3

    NascentOxygen

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    Staff: Mentor

    Some of the electrical energy is "used up" in splitting the water. The excess of electrical input heats the vessel contents: the gas mixture and the remaining liquid, and these both expand.
     
  5. Jul 13, 2015 #4
    Does 3.96 describe the gaseous space of the container? This would make sense to me except that the initial liquid volume was 36mL. I don't understand why it's only 3.96mL.

    0.02798g is produced after 1 min. I used this value to solve for total temperature of the system and used that value to solve for the change in volume. Did I use the right equation to solve for change in volume?
     
  6. Jul 13, 2015 #5
    Thank you for the clarification! I used the ideal gas law to solve for the change in volume:
    upload_2015-7-13_16-9-22.png

    Since pressure, moles, and gas constant were the same, I was left with this equation:
    upload_2015-7-13_16-11-38.png

    Was this the right equation to use?
     

    Attached Files:

  7. Jul 13, 2015 #6
    Yes 3.96 is the volume of gas in the container. We differ perhaps because we used different atomic wgts?

    No gas law is needed. The converted water gives the additional volume. .02798 ml because the density of water is 1 gm/ml. Temperature and pressure have little affect.
     
  8. Jul 13, 2015 #7

    Borek

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    Staff: Mentor

    I told you in another thread this is self contradicting. You can't have a closed container and a standard pressure, as the pressure changes when the gas is produced. You either have a constant volume, or constant pressure. Each time you post another iteration of this question you word it differently, but never precisely enough, so each time people help you solve slightly different problem - not knowing what the problem really is.

    This is a waste of time.
     
  9. Jul 13, 2015 #8
    I apologize. Still very confused about how to create and solve these types of equations. I don't know how to solve for the final volume and pressure during electrolysis when I only know the initial temperature and volume.

    You've been very patient and incredibly helpful. I apologize if it appears that I'm asking the same question over and over again. But honestly, I'm just using the same values to make it easier for me to understand. I'm basically trying to learn how to calculate internal energy under nonstandard conditions during electrolysis when only the amps, voltage, moles of the substance, initial temperature and volume of the container are known. You've helped me calculate the final temperature of the system during electrolysis. Now I'm trying to learn how to solve for the change in pressure and volume.
     
  10. Jul 13, 2015 #9
    Actually, with a rigid body container the volume should be fixed and it's pressure that I'd really be solving for isn't it?
     
  11. Jul 18, 2015 #10

    Borek

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    Staff: Mentor

    Yes.
     
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