# Electrolysis of water and charges help

• gymstar
In summary, the question involves the electrolysis of water to produce 11.2L of oxygen at STP. To find the required charge, the molar volume of oxygen at STP is used to determine the number of moles of oxygen produced. Then, the number of electrons required for each molecule of oxygen is calculated and converted to coulombs. For the second question, the formula t = Q/I is used to find the time required for a current of 0.05A to produce the given amount of oxygen.
gymstar
Hi i am confused about on how to solve this question.

The question asks: By the electrolysis of water, 11.2L oxygen at STP was prepared. a) What charge was required? b) if a current of 0.05A was used, how long did it take?

I have done the question but i think i did it wrong.

2H2O(l) ---> O2(g) + 4H +(aq) + 4e- (oxidation)
4H +(aq) + 4e- ---> 2H2O(g) (reduction)

a) 1mol/22.4L = 0.0446
0.0446* 96500= 4303.9 4.3* 10^3

b) 0.5A = 4.3* 10^3/t
4.3*10^3/0.5A
t = 8607.8s

Is this completely wrong?

First your thread title is misleading since this is a question about electrochemistry. Second I'm not sure what is meant by "What charge was required?" Could you mean what potential (voltage) is required ? If so simply solve the redox reaction for the potential. Finally for b) simply convert the O2 to moles and convert the .05 A to moles of electrons/sec and due the math :)

I think your full reaction should be $$2H_2O (l) \longrightarrow O_2 (g) + 2H_2 (g)$$

Your standard half reactions are $$O_2 (g) + 4 H^+ (aq) + 4 e^- \longrightarrow 2H_2O \ E^0 = +1.23 V$$ and
$$2H_2O (l) + 2 e^- \longrightarrow H_2 (g) + 2 OH^- (aq)\ E^0 = -0.83 V$$

I got these reduction potentions from http://www.jesuitnola.org/upload/clark/Refs/red_pot.htm Of course, during electrolysis the first reaction occurs in reverse. So what do you think the total potential would be?

Edit: Actually you don't need to bother about the potentials.

a) You have 11.2 L of O2 at 1 atm and 273 K. How many moles of O2 is this? Hint: PV = nRT. Now how many electorns are required to reduce 1 molecule of O2 ? 1 mol of O2? How many coulombs are there in x moles of electrons ?

b) This one should be easy once you have (a) 1 ampere = 1 coulomb / second. You have coulombs and amps. Solve for seconds.

Last edited by a moderator:
The question asks: By the electrolysis of water, 11.2L oxygen at STP was prepared. a) What charge was required? b) if a current of 0.05A was used, how long did it take?

a)at STP 1mole/22.4L (I think), convert to moles

-using the half reaction equation, find out the mole of electrons required for every 1 mole of oxygen. Convert to moles of electrons using factor labeling.

-convert this mole value to avogadros number. From this convert to charge using the charge of one electron.

-

Hi... I am working on this one too and would appreciate it if anyone could take the time to check my answers.

By the electrolysis of water, 11.2L oxygen at STP was prepared.
a) What charge was required?
b) if a current of 0.05A was used, how long did it take?

A) the text lists the following reaction
H2O <-> 2H+ + .5O2 + 2e- -0.82
since half a mol of O2 is produced the charge was the listed (-0.82)value in the equation as the equation is for half a mol of O2.

B)Q=It or t = Q/I
t= .5 mol / .5 amp = 1 second
(not too sure on this one for the value of Q - the quantity of electric charge transferred through a cell)

Maybe I am missing something but I did this question very differently than Gymstar and Yellowduck:

a) The molar volume of any gas at STP is 22.4 L
Therefore 11.2 L of oxygen at STP = 1/2 mole
1 molecule of oxygen requires 4 electrons (each O requires 2 electrons $$O^2^-$$and oxygen is diatomic $$O_2$$)
Therefore 1/2 mole of oxygen requires 4*1/2*1 mole of electrons
1 mole of electrons = 9.64*10^4 C

(4)(1/2)(9.64*10^4C) = 1.93*10^5 C

b) t = Q/I

t = (1.93*10^5 C) / 0.5 A = 386000 seconds or 107.2 hours

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## 1. What is electrolysis of water?

Electrolysis of water is a process in which an electric current is passed through water to separate it into its two components: hydrogen and oxygen gas. This is achieved by breaking the bonds between the hydrogen and oxygen atoms in water molecule, resulting in the production of hydrogen gas at the cathode and oxygen gas at the anode.

## 2. Can water conduct electricity?

Pure water does not conduct electricity. This is because it does not contain any free ions that can carry electric charge. However, when an electrolyte (such as a salt) is added to water, it becomes able to conduct electricity as the ions from the electrolyte dissociate and can carry electric charge.

## 3. What is the purpose of using an electrolyte in electrolysis of water?

The electrolyte is necessary in electrolysis of water because it provides ions that can carry electric charge. Without an electrolyte, the water would not be able to conduct electricity and the process of electrolysis would not be possible.

## 4. How does the charge of the electrodes affect the products of electrolysis?

The charge of the electrodes plays a crucial role in determining the products of electrolysis. The cathode (negative electrode) attracts positively charged ions, while the anode (positive electrode) attracts negatively charged ions. This means that the products of electrolysis will be different depending on which electrode they are attracted to.

## 5. What are the applications of electrolysis of water?

Electrolysis of water has several important applications, including the production of hydrogen gas for fuel, the production of oxygen gas for medical and industrial purposes, and the purification of water. It is also an important process in the production of metals and other chemical compounds.

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