# Need help with oxidation question

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#4

Al2(SO4)3

(SO4)2- has 2- charge

if Al in aluminium sulfate has ON x

then 2x + 3(-2) = 0 giving x = +3

(SO4)2- has 2- charge

if Al in aluminium sulfate has ON x

then 2x + 3(-2) = 0 giving x = +3

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(Original post by

Al2(SO4)3

(SO4)2- has 2- charge

if Al in aluminium sulfate has ON x

then 2x + 3(-2) = 0 giving x = +3

**golgiapparatus31**)Al2(SO4)3

(SO4)2- has 2- charge

if Al in aluminium sulfate has ON x

then 2x + 3(-2) = 0 giving x = +3

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**BlueDiamondzz**)

I don’t understand the 3rd and 4th line of what you wrote can you please explain to me further

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(Original post by

**BlueDiamondzz**)
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#8

Al can't have oxidation number 0 because then the charges don't balance in Al2(SO4)3

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(Original post by

Al can't have oxidation number 0 because then the charges don't balance in Al2(SO4)3

**golgiapparatus31**)Al can't have oxidation number 0 because then the charges don't balance in Al2(SO4)3

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(Original post by

How do you calculate the charger of sulphur (sorry I’m really dumb). And what’s ON x?

**BlueDiamondzz**)How do you calculate the charger of sulphur (sorry I’m really dumb). And what’s ON x?

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#11

**BlueDiamondzz**)

How do you calculate the charger of sulphur (sorry I’m really dumb). And what’s ON x?

The sulfur in it has an oxidation number (which I called ON) +6

I called the ON of aluminium x because it's unknown, then I solved for it. Did your teacher not explain this topic??

(Original post by

If we’re turning sulphur and oxygen it into an ion and there’s the remaining 1 do we transfer it to the aluminium to make it Al3 instead if Al2?

**BlueDiamondzz**)If we’re turning sulphur and oxygen it into an ion and there’s the remaining 1 do we transfer it to the aluminium to make it Al3 instead if Al2?

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(Original post by

The sulfate ion SO4^2- has a charge of 2-

The sulfur in it has an oxidation number (which I called ON) +6

I called the ON of aluminium x because it's unknown, then I solved for it. Did your teacher not explain this topic??

Sorry I don't understand..

**golgiapparatus31**)The sulfate ion SO4^2- has a charge of 2-

The sulfur in it has an oxidation number (which I called ON) +6

I called the ON of aluminium x because it's unknown, then I solved for it. Did your teacher not explain this topic??

Sorry I don't understand..

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Our teacher told us we turn it into an ion or something like that. I don’t remember and I thought I knew this topic but it turns out I don’t. I currently have no contact with the teacher so I can’t ask for help 😕

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#14

(Original post by

I understand now that x is 2x because aluminium is Al2. But I don’t understand why you did 3 x (-2). Is the -2 from the oxygen? And the 3 from the 3 part of the bracket of (SO4)3? Do you mind explaining it in steps and tell me if we need to turn it into an ion? Or maybe present your working out in a visual way. Sorry for bothering you. I have learning difficulties so I know it’s tough to explain things to me

**BlueDiamondzz**)I understand now that x is 2x because aluminium is Al2. But I don’t understand why you did 3 x (-2). Is the -2 from the oxygen? And the 3 from the 3 part of the bracket of (SO4)3? Do you mind explaining it in steps and tell me if we need to turn it into an ion? Or maybe present your working out in a visual way. Sorry for bothering you. I have learning difficulties so I know it’s tough to explain things to me

Al

_{2}(SO

_{4})

_{3}

The way I'm doing it is considering the Al and SO4 units separately

I know that SO

_{4}

^{2-}has 2- charge.

The oxidation number of aluminium (I call it x because I need to calculate it, it is unknown) is equal to its charge

The overall charge is zero.

If I add charges of 2 Al(x+) units and 3 SO4(2-) units I get 0.

so 2x + 3(-2) = 0

2x = 6

x = +3

so the oxidation number is +3

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I would also like to know the oxidation number for sulphur? I thought we times by the bracket for Al2(SO4)3 so I thought we would times 4 by 3

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(Original post by

I would also like to know the oxidation number for sulphur? I thought we times by the bracket for Al2(SO4)3 so I thought we would times 4 by 3

**BlueDiamondzz**)I would also like to know the oxidation number for sulphur? I thought we times by the bracket for Al2(SO4)3 so I thought we would times 4 by 3

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#17

To calculate oxidation number of sulfur in sulfate (SO4)2-:

use the fact that you know the oxidation number of oxygen as -2

then call the oxidation number of S x

x + 4(-2) = -2

so x = +6

use the fact that you know the oxidation number of oxygen as -2

then call the oxidation number of S x

x + 4(-2) = -2

so x = +6

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(Original post by

To calculate oxidation number of sulfur in sulfate (SO4)2-:

use the fact that you know the oxidation number of oxygen as -2

then call the oxidation number of S x

x + 4(-2) = -2

so x = +6

**golgiapparatus31**)To calculate oxidation number of sulfur in sulfate (SO4)2-:

use the fact that you know the oxidation number of oxygen as -2

then call the oxidation number of S x

x + 4(-2) = -2

so x = +6

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#19

(Original post by

I’m feeling confused again (sorry). I know that Al2(SO4)3 and we turn (SO4)3 into (SO4)2- then what happens to the remaining 1 or does that disappear because it’s a charge now?

**BlueDiamondzz**)I’m feeling confused again (sorry). I know that Al2(SO4)3 and we turn (SO4)3 into (SO4)2- then what happens to the remaining 1 or does that disappear because it’s a charge now?

Al2(SO4)3 is a shorthand way of writing [Al

^{3+}]

_{2}[SO

_{4}

^{2-}]

_{3}

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**BlueDiamondzz**)

I’m feeling confused again (sorry). I know that Al2(SO4)3 and we turn (SO4)3 into (SO4)2- then what happens to the remaining 1 or does that disappear because it’s a charge now?

Is this right?

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