Mutleybm1996
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How would I write half equations for the H+ in (C)?

Thanks


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charco
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How would I write half equations for the H+ in (C)?

Thanks


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You don't...

... it's not one of the oxidised or reduced species.

It's used to 'mop up' the oxygen from the manganate(VII) ion.

In this equation the manganate(VII) ion is reduced and the chloride ions are oxidised.
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Mutleybm1996
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(Original post by charco)
You don't...

... it's not one of the oxidised or reduced species.

It's used to 'mop up' the oxygen from the manganate(VII) ion.

In this equation the manganate(VII) ion is reduced and the chloride ions are oxidised.
Thank you!
I have to assign an oxidation number to Na, would this be +1? What about Na^+? Isn't this +1 too?


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charco
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(Original post by Mutleybm1996)
Thank you!
I have to assign an oxidation number to Na, would this be +1? What about Na^+? Isn't this +1 too?


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Elements are always 0
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cupsoftea
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(Original post by Mutleybm1996)
Thank you!
I have to assign an oxidation number to Na, would this be +1? What about Na^+? Isn't this +1 too?


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Na oxidation number as an element? 0
Na+ as an ion +1
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Mutleybm1996
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(Original post by cupsoftea)
Na oxidation number as an element? 0
Na+ as an ion +1
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(Original post by Mutleybm1996)
So are there correct? Name:  ImageUploadedByStudent Room1412101424.209850.jpg
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For As^3- , where did you get +8 from?

Also Br-, why do you think -2?

Answers for both are their respective charges, so -3, and -1. Rule: The oxidation number of a monoatomic ion equals the charge of the ion.

The rest are correct
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Mutleybm1996
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(Original post by cupsoftea)
For As^3- , where did you get +8 from?

Also Br-, why do you think -2?

Answers for both are their respective charges, so -3, and -1. Rule: The oxidation number of a monoatomic ion equals the charge of the ion.

The rest are correct
Thank you! A minor slip up! *guilty as charged!*

Could I possibly run these past you? I think I have most correct! We were taught this rather hastily and the textbook isn't the clearest.
If it would take too much time then it's no problem!

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cupsoftea
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(Original post by Mutleybm1996)
Thank you! A minor slip up! *guilty as charged!*

Could I possibly run these past you? I think I have most correct! We were taught this rather hastily and the textbook isn't the clearest.
If it would take too much time then it's no problem!

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Had a scan of them, can see one (maybe more) are wrong from a quick glance.
It's not the clearest which parts are from the question - could you tell me which parts you've extracted from the question itself? Then I'll explain how I would go through the question from the information given/be able to tell what you've done wrong/right better
The key is to have all charges and elements balanced on either side of the final equation - that is usually a good check for whether it is correct or not, if its not balanced something has gone wrong.
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Mutleybm1996
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(Original post by cupsoftea)
Had a scan of them, can see one (maybe more) are wrong from a quick glance.
It's not the clearest which parts are from the question - could you tell me which parts you've extracted from the question itself? Then I'll explain how I would go through the question from the information given/be able to tell what you've done wrong/right better
The key is to have all charges and elements balanced on either side of the final equation - that is usually a good check for whether it is correct or not, if its not balanced something has gone wrong.
Top line=equation
Next two=my attempts at half equations
Final line=combining the half equations


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GDN
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you have to construct a reduction half equation and an oxidation half equation then eliminate the electrons - each half equation and the final balanced equation must be balanced both in terms of atoms (a mass balance) and in terms of charge - afraid to say all of them are incorrect
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(Original post by Mutleybm1996)
Top line=equation
Next two=my attempts at half equations
Final line=combining the half equations


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Ok so half equations for 1st one

Fe2+ --> Fe3+ + e- Oxidation

Cr2O72- + 6e- + 14H+ --> 2Cr3+ + 7H2O Reduction

First one with iron is a simple loss of an electron, second one with Cr is gain of electrons then balanced with H+ and H2O

Notice all charges balance, for the second half equation on the left hand side you have 2- + 6- + 14 = +6 and right hand side 2x+3 = +6, so both sides have charge of +6 (vital they are equal otherwise something has gone wrong in your working)

Then combining, multiply the Fe's by 6, as Cr2O72- needs 6e- to be reduced to two lots of Cr3+. so an electron from each of Fe2+

Overall redox equation

Cr2O72- + 6Fe2+ + 14H+ --> 2Cr3+ + 7H2O + 6FE3+

Now check charges balance, +24 on either side, so they do. Also equal amounts of each element on each side, thus you know it is correct
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(Original post by GDN)
you have to construct a reduction half equation and an oxidation half equation then eliminate the electrons - each half equation and the final balanced equation must be balanced both in terms of atoms (a mass balance) and in terms of charge - afraid to say all of them are incorrect
Could you explain it? I thought the 1st one was balanced both in charges and atoms?
I'm confused, sorry.
Can you explain further where I went wrong with them?


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cupsoftea
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(Original post by Mutleybm1996)
Top line=equation
Next two=my attempts at half equations
Final line=combining the half equations


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I'll explain the last one too

MnO4- is being reduced to Mn2+

MnO4-, Mn has an oxidation state of +7

thus to go from MnO4- to Mn2+, an addition of 5e- must occur

MnO4- + 5e- --> Mn2+

This is *incomplete*. This must be balanced!! There is both unequal charge and elements on either side. To balance, we use water and H+

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

Notice how each side overall charge of 2+

Iron is being oxidised, 5 lots to account for the 5e- the Manganese needs above

5Fe2+ --> 5Fe3+ + 5e-

Then simple case of adding the two half equations together


Any questions ask, if you think you understood it, have a go at one of the other questions using these principles and i'll have a look, remember to check if its right, everything must be balanced (elements and charge wise)
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Mutleybm1996
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(Original post by cupsoftea)
I'll explain the last one too

MnO4- is being reduced to Mn2+

MnO4-, Mn has an oxidation state of +7

thus to go from MnO4- to Mn2+, an addition of 5e- must occur

MnO4- + 5e- --> Mn2+

This is *incomplete*. This must be balanced!! There is both unequal charge and elements on either side. To balance, we use water and H+

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

Notice how each side overall charge of 2+

Iron is being oxidised, 5 lots to account for the 5e- the Manganese needs above

5Fe2+ --> 5Fe3+ + 5e-

Then simple case of adding the two half equations together


Any questions ask, if you think you understood it, have a go at one of the other questions using these principles and i'll have a look, remember to check if its right, everything must be balanced (elements and charge wise)
So is every one wrong? On both sides?

Thanks so far! I may have more questions later


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charco
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(Original post by Mutleybm1996)
Could you explain it? I thought the 1st one was balanced both in charges and atoms?
I'm confused, sorry.
Can you explain further where I went wrong with them?


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Try these interactive explanations. (click on the highlighted section to advance)

Redox equations 1
Redox equations 2
Redox equations 3
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cupsoftea
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(Original post by Mutleybm1996)
So is every one wrong? On both sides?

Thanks so far! I may have more questions later


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Yes

You might want to have a look online at videos on how to do these

In the final question, MnO4- is the species being reduced, but it isn't featured in your final equation? same for the SO3 in earlier questions.

MnO4- has oxidation number +7, but it isn't Mn7+, you treat it as MnO4- when writing the equations. The only extra thing to consider for these species compared to a simple Fe2+/F3+ reaction, is balancing the equations with H+ and H2O.

So for this example, MnO4- has 4 oxygens, so that is 4H2O needed on the RHS

Breakdown of process for this half equation:

Step 1, addition of 5e- to reduce 7+ to 2+
MnO4- + 5e- --> Mn2+

Step 2: balance the oxygen by adding apprpriate amount of water (4)

MnO4- + 5e- --> Mn2+ + 4H2O

Step 3: balance the hydrogen (which has appeared in the water from step 2)

As you've just added 4 H2O on the RHS, you have 8Hydrogens there now, so you need to also add 8 Hydrogens on the left, thus add 8H+ to the LHS

MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O

that is how that half equation comes to be
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(Original post by cupsoftea)
Yes

You might want to have a look online at videos on how to do these

In the final question, MnO4- is the species being reduced, but it isn't featured in your final equation? same for the SO3 in earlier questions.

MnO4- has oxidation number +7, but it isn't Mn7+, you treat it as MnO4- when writing the equations. The only extra thing to consider for these species compared to a simple Fe2+/F3+ reaction, is balancing the equations with H+ and H2O.

So for this example, MnO4- has 4 oxygens, so that is 4H2O needed on the RHS

Breakdown of process for this half equation:

Step 1, addition of 5e- to reduce 7+ to 2+
MnO4- + 5e- --> Mn2+

Step 2: balance the oxygen by adding apprpriate amount of water (4)

MnO4- + 5e- --> Mn2+ + 4H2O

Step 3: balance the hydrogen (which has appeared in the water from step 2)

As you've just added 4 H2O on the RHS, you have 8Hydrogens there now, so you need to also add 8 Hydrogens on the left, thus add 8H+ to the LHS

MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O

that is how that half equation comes to be
Thank you once again! Can you have a quick look and tell me which of a-m are correct so i don't have to re-do them?
Thanks again, sorry! :/


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cupsoftea
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(Original post by Mutleybm1996)
Thank you once again! Can you have a quick look and tell me which of a-m are correct so i don't have to re-do them?
Thanks again, sorry! :/


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They're all wrong

You can tell because the species in the question equations must be in your final equation - you have changed the species (for example the MnO4- you've said is Mn7+)

If you understand the explanations I have given for the first and last question you should be able to recognize none of the others can be right.

A further hint is that Cr2O7- is being reduced to Cr3+ not Cr2+, that should allow you to balance them properly in both elements and charge as you currently have them down as going to Cr2+

Follow the steps above for balancing H+ and H2O and you should be able to fairly quickly correct the others. I'm happy to recheck your new answers
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Mutleybm1996
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(Original post by cupsoftea)
They're all wrong

You can tell because the species in the question equations must be in your final equation - you have changed the species (for example the MnO4- you've said is Mn7+)

If you understand the explanations I have given for the first and last question you should be able to recognize none of the others can be right.

A further hint is that Cr2O7- is being reduced to Cr3+ not Cr2+, that should allow you to balance them properly in both elements and charge as you currently have them down as going to Cr2+

Follow the steps above for balancing H+ and H2O and you should be able to fairly quickly correct the others. I'm happy to recheck your new answers
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Hi again, how would i balance this?


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