# Balancing redox equations help

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#1
Q) Balance:
Cl2 + NaOH -----> NaClO3 + NaCl + H2O

A) Cl2 +2NaOH +2H2O ------> NaClO3 +NaCl + H2O + 3H+ + 5e-

Thanks
0
2 years ago
#2
(Original post by anactualmess)
Q) Balance:
Cl2 + NaOH -----> NaClO3 + NaCl + H2O

A) Cl2 +2NaOH +2H2O ------> NaCl + H2O + 3H+ + 5e-

Thanks
no
0
#3
(Original post by charco)
no
0
2 years ago
#4
(Original post by anactualmess)
Your answer cannot possibly be correct because it does not even contain the same species ...

Step 1: identify the species that is oxidised and generate a half-equation using only water and hydroxide ions to balance the atoms and add in electrons to balance the charge

Step 2: do the same for the reduced species

Step 3: Multiply your two equations by suitable integers to equalise the electrons

Step 4: add the half-equations together.

Step 5: Gather terms and cancel down if possible.

Step 6: Add the balancing ions (spectator ions)
0
#5
Oh sorry I typed it up wrong, I meant to type :

A) Cl2 +2NaOH +2H2O ------> NaClO3 +NaCl + H2O + 3H+ + 5e-

(Original post by charco)
Your answer cannot possibly be correct because it does not even contain the same species ...

Step 1: identify the species that is oxidised and generate a half-equation using only water and hydroxide ions to balance the atoms and add in electrons to balance the charge

Step 2: do the same for the reduced species

Step 3: Multiply your two equations by suitable integers to equalise the electrons

Step 4: add the half-equations together.

Step 5: Gather terms and cancel down if possible.

Step 6: Add the balancing ions (spectator ions)
0
2 years ago
#6
(Original post by anactualmess)
Oh sorry I typed it up wrong, I meant to type :

A) Cl2 +2NaOH +2H2O ------> NaClO3 +NaCl + H2O + 3H+ + 5e-
A balanced equation has to have the same number and type of atoms on either side and the same electrical charge on etiher side.

You can't just add in random species that are not in the question.

Follow the steps I gave you ...
0
2 years ago
#7
Q) Balance:
Cl2 + NaOH -----> NaClO3 + NaCl + H2O

Hi, let me show you a tactic to do balancing step by step, yeah?

1) First, because there are 2 nitrogens on the right side, you will need at least 2NaOH on left side.

Cl2 + 2NaOH -----> NaClO3 + NaCl + H2O

2) However, if there are 2NaOH on left, it will not be enough, because this would give 2 oxygens on left but still 4 Os on right,
3) If you therefore make it 4NaOH on left, AND you will need to make 3NaCl on right (you can't have more NaClO3 as it will make too many Os on the right) as below
(Cl2 + 4NaOH -----> 2NaClO3 + 2NaCl + H2O)[not correct]

4) Now (you will then need 2Cl2 as well on left cos there are 4 Cl-s on right (still with me?)
Now you have:

2 Cl2 + 4NaOH -----> NaClO3 + 3NaCl + 2H2O

Is this balanced? check all the elements ONE BY ONE! What do u say?............................ ..........................
................................ ...........

....(YES well done: it is balanced.

Are you happy with the step y step logic? That is how you approach any balancing of equations. (I am a biologist/medic so chemistry is not my subject and I actually have not come across this equation BUT BECAUSE WE HAVE MANAGED TO BALANCE IT, IT SHOULD BE POSSIBLE!))

HANG ON !! Now there are 4 Os on left BUT 5 on R - CHECK THIS Q - I think the formula of sodium hypochlorite is actually NaClO (without a 3)

THE REACTION CANNOT OCCUR!!!

So sorry!

M
0
2 years ago
#8
(Original post by anactualmess)
Q) Balance:
Cl2 + NaOH -----> NaClO3 + NaCl + H2O

A) Cl2 +2NaOH +2H2O ------> NaClO3 +NaCl + H2O + 3H+ + 5e-

Thanks
You've confused balancing equations with half-equations.
0
2 years ago
#9
(Original post by macpatelgh)
Q) Balance:
Cl2 + NaOH -----> NaClO3 + NaCl + H2O

Hi, let me show you a tactic to do balancing step by step, yeah?

1) First, because there are 2 nitrogens on the right side, you will need at least 2NaOH on left side.

Cl2 + 2NaOH -----> NaClO3 + NaCl + H2O

2) However, if there are 2NaOH on left, it will not be enough, because this would give 2 oxygens on left but still 4 Os on right,
3) If you therefore make it 4NaOH on left, AND you will need to make 3NaCl on right (you can't have more NaClO3 as it will make too many Os on the right) as below
(Cl2 + 4NaOH -----> 2NaClO3 + 2NaCl + H2O)[not correct]

4) Now (you will then need 2Cl2 as well on left cos there are 4 Cl-s on right (still with me?)
Now you have:

2 Cl2 + 4NaOH -----> NaClO3 + 3NaCl + 2H2O

Is this balanced? check all the elements ONE BY ONE! What do u say?............................ ..........................
................................ ...........

....YES well done: it is balanced.

Are you happy with the step y step logic? That is how you approach any balancing of equations. (I am a biologist/medic so chemistry is not my subject and I actually have not come across this equation BUT BECAUSE WE HAVE MANAGED TO BALANCE IT, IT SHOULD BE POSSIBLE!)

M
this is actually the reaction of halogens with hot alkali. There's also the cold alkali one where you would get NaClO instead of NaClO3
0
2 years ago
#10
(Original post by macpatelgh)

Now you have:

2 Cl2 + 4NaOH -----> NaClO3 + 3NaCl + 2H2O

Is this balanced?
No, it's not balanced.

Check out the oxygen
0
2 years ago
#11
(Original post by charco)
No, it's not balanced.

Check out the oxygen
Correct I just edited it - perhaps you can help out the young student!
0
2 years ago
#12
(Original post by RickHendricks)
this is actually the reaction of hydrogen halides with hot alkali. There's also the cold alkali one where you would get NaClO instead of NaClO3
Hi, thanks for your input - but surely a hydrogen halide should be e.g. HCl NOT Cl2??
0
2 years ago
#13
(Original post by macpatelgh)
Hi, thanks for your input - but surely a hydrogen halide should be e.g. HCl NOT Cl2??
oh my mistake.

Meant to say halogens.

lemme edit it
1
#14
(Original post by charco)
Your answer cannot possibly be correct because it does not even contain the same species ...

Step 1: identify the species that is oxidised and generate a half-equation using only water and hydroxide ions to balance the atoms and add in electrons to balance the charge

Step 2: do the same for the reduced species

Step 3: Multiply your two equations by suitable integers to equalise the electrons

Step 4: add the half-equations together.

Step 5: Gather terms and cancel down if possible.

Step 6: Add the balancing ions (spectator ions)
Turns out I was mistaking the equation for a half equation lol.

Following your steps (thank you for that!) I managed to establish the two half equations and have come to the answer:

3Cl2 + 6NaOH ------> NaClO3 + 5NaCl + 3H2O

I just have one more question, when balancing half equations, I know you use hydroxide ions to balance hydrogen when in basic solution and hydrogen ions to balance hydrogen when in acidic solution, so how do you know if your'e dealing with an acidic or basic solution? (We haven't covered this in lessons)
0
#15
(Original post by RickHendricks)
You've confused balancing equations with half-equations.
Thank you for pointing that out!
0
2 years ago
#16
cl2 + 2NaOH -------> NaCL + NaClO + H2O
0
2 years ago
#17
(Original post by anactualmess)
Turns out I was mistaking the equation for a half equation lol.

Following your steps (thank you for that!) I managed to establish the two half equations and have come to the answer:

3Cl2 + 6NaOH ------> NaClO3 + 5NaClO3 + 5NaCl + 3H2O

I just have one more question, when balancing half equations, I know you use hydroxide ions to balance hydrogen when in basic solution and hydrogen ions to balance hydrogen when in acidic solution, so how do you know if your'e dealing with an acidic or basic solution? (We haven't covered this in lessons)
You haven't quite arrived at the correct answer. your atoms are not balanced.

I haven't seen your working so I don't know where your error is, but if you followed the steps I gave you it must be a maths error.

Half-equation 1: Cl2 + 2e --> 2Cl-
Half-equation 2: Cl2 + 12OH- --> 2ClO3- + 6H2O + 10e

Multiply through by 5 in equation 1 and add together:

Half-equation 1: 5Cl2 + 10e --> 10Cl-
Half-equation 2: Cl2 + 12OH- --> 2ClO3- + 6H2O + 10e
5Cl2 + Cl2 + 12OH- --> 10Cl- + 2ClO3- + 6H2O

gather terms:

6Cl2 + 12OH- --> 10Cl- + 2ClO3- + 6H2O

divide through by 2:

3Cl2 + 6OH- --> 5Cl- + ClO3- + 3H2O

Now add in the spectator ions (sodium):

3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O

It's not an easy reaction to deal with as it involves redox (disporoportionation) under basic conditions. But following a stepwise procedure you should always arrive at the correct answer.
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