Water electrolysis - solving for final volume of the system

• HelloCthulhu
In summary, after 1 min of electrolysis, 0.02798 grams of gas was produced. The temperature of the system rose from 20°C to 99.7°C.
HelloCthulhu

Homework Statement

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?

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

Q=c x m x ΔT

Vi/Ti = Vf/Tf

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

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:
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.

gleem said:
initially it is about 3.96 ml and after electolysis is will be slightly greater due to the change of liquid to gaseous products

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.

gleem said:
So how much water in converted?

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?

NascentOxygen said:
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.

Thank you for the clarification! I used the ideal gas law to solve for the change in volume:

Since pressure, moles, and gas constant were the same, I was left with this equation:

Attachments

626 bytes · Views: 490
HelloCthulhu said:
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.

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.

HelloCthulhu said:
closed 40mL container for 1 min at standard external/internal pressure

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.

Borek said:
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.

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.

Borek said:
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.

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.

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?

HelloCthulhu said:
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?

Yes.

1. What is water electrolysis?

Water electrolysis is a process in which an electric current is passed through water to break it down into its component parts, hydrogen and oxygen. This process is commonly used to produce hydrogen gas for industrial and energy storage purposes.

2. How does water electrolysis work?

During electrolysis, water molecules are split into positively charged hydrogen ions (protons) and negatively charged hydroxide ions. The electric current causes these ions to move towards the respective electrodes, where they combine to form hydrogen gas at the negative electrode and oxygen gas at the positive electrode.

3. What factors affect the final volume of the system in water electrolysis?

The final volume of the system in water electrolysis is affected by several factors, including the amount of water used, the strength of the electric current, the duration of the electrolysis process, and the type of electrodes used.

4. How can the final volume of the system be calculated in water electrolysis?

The final volume of the system can be calculated by multiplying the number of moles of hydrogen gas produced by the molar volume of gas at the given temperature and pressure. This can be determined by using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

5. What are the main applications of water electrolysis?

Water electrolysis has several applications, including the production of hydrogen gas for fuel cells and other industrial processes, energy storage in the form of hydrogen gas, and the production of oxygen gas for medical, industrial, and scientific purposes. It can also be used in the purification of water and in the production of certain chemicals.

• Biology and Chemistry Homework Help
Replies
26
Views
4K
• Biology and Chemistry Homework Help
Replies
3
Views
1K
• Biology and Chemistry Homework Help
Replies
4
Views
2K
• Chemistry
Replies
4
Views
2K
• Biology and Chemistry Homework Help
Replies
8
Views
6K
• Biology and Chemistry Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
871
• Biology and Chemistry Homework Help
Replies
4
Views
3K
• Biology and Chemistry Homework Help
Replies
11
Views
25K
• Biology and Chemistry Homework Help
Replies
1
Views
6K