redox question

It doesn't.

Reply 2

OP

Kinkerz

It doesn't.

teacher hand-wrote some answers..maybe she did it wrong.

how would u attempt this quest?

Reply 3

joshed

8

That equation doesn't balance. Where has the N gone ? Think about that to help you work out oxidation states first :smile:

Reply 4

LaraArmstrong

O2 is a molecule, therefore doesnt have a charge.

Reply 5

OP

joshed

That equation doesn't balance. Where has the N gone ? Think about that to help you work out oxidation states first :smile:

aah typo...sorted that out nw now...could someone please help me now. :smile:

Reply 6

Well for one, the equation isn't even right. Where's the nitrogen gone? You can't have elements in the reactants and not in the products.

Ba(NO₃)₂ ---> BaO + N₂O₅

No?

---

Oxygen usually has an oxidation state of -2, and that's how it remains in this I reckon.

Reply 7

OP

Kinkerz

Well for one, the equation isn't even right. Where's the nitrogen gone? You can't have elements in the reactants and not in the products.

Ba(NO₃)₂ ---> BaO + N₂O₅

No?

---

Oxygen usually has an oxidation state of -2, and that's how it remains in this I reckon.

sortred the equation mess out..i frgt to include it..whilst typing..tizz included now..but still don't gerrit. :frown:

Reply 8

FoOtYdUdE

aah typo...sorted that out nw now...could someone please help me now. :smile:

The oxidation state of oxygen isn't changing really.

It's nitrogen that has the variable oxidation state here.

Reply 9

OP

Kinkerz

The oxidation state of oxygen isn't changing really.

It's nitrogen that has the variable oxidation state here.

well i'll tell u what's bugging me. it's the 2Ba (NO3)2

i dunno how to sort that out....the products side is easy (i.e right) ... it's the left hand side which i cant sort out. am i making sense? :confused:

Reply 10

LearningMath

14

No, you're not. You first need to ask a coherent question.

Reply 11

FoOtYdUdE

well i'll tell u what's bugging me. it's the 2Ba (NO3)2

i dunno how to sort that out....the products side is easy (i.e right) ... it's the left hand side which i cant sort out. am i making sense? :confused:

You're not being that clear on what you're aiming to do.

Ba(NO₃)₂

Oxidation states:
Ba = +2
O = -2

Therefore you can work out the oxidation state of N...

Reply 12

OP

ok, so it's 2 Ba(NO3)2

shall i ignore this 2??

if i ignore it, i'm left with:

Ba(NO3)2

for O : (6 x -2 = -12)
for Ba : (= + 2)

so -12 + 2 = -10

:confused:

how do i work out N?

N2 = - 10 ???

N= -5

?? :confused:

Reply 13

bb193

13

What reaction are we talking about here? Is it thermal decomposition?

2Ba(NO3)2 + heat → 2BaO + 4NO2 + O2 ?

Or something else?

Reply 14

OP

THE QUESTION:

Barium nitrate decomposes when heated to make barium oxide, nitrogen dioxide and oxygen.

2Ba (NO3)2 >>> 2BaO + O2 + 4NO2

USE oxidation states to explain why this decomposition reaction involves both oxidation and reduction.

Reply 15

FoOtYdUdE

ok, so it's 2 Ba(NO3)2

shall i ignore this 2??

if i ignore it, i'm left with:

Ba(NO3)2

for O : (6 x -2 = -12)
for Ba : (= + 2)

so -12 + 2 = -10

:confused:

how do i work out N?

N2 = - 10 ???

N= -5

?? :confused:

It's not going to be -5 if you're trying to make -10 upto 0 is it? Think about it.

If you've got 2N to make -10 upto 0, then each N must be +5.

Reply 16

OP

Kinkerz

It's not going to be -5 if you're trying to make -10 upto 0 is it? Think about it.

If you've got 2N to make -10 upto 0, then each N must be +5.

damn, silly me...yes
ok so that's the nitrogen sorted....

what about the oxygen...miss wrote -2...dyu think she made a typo...and it was -12 ??? for O

Reply 17

FoOtYdUdE

damn, silly me...yes
ok so that's the nitrogen sorted....

what about the oxygen...miss wrote -2...dyu think she made a typo...and it was -12 ??? for O

Each O is -2. The total O make up -12 as there are 6 of them.

Nitrogen is going from +5 in Ba(NO₃)₂ to +4 in NO₂. Therefore you can safely assume that is has gained an electron and hence gone through a reduction.

Reply 18

bb193

13

Many things can have different oxidation states. But oxygen is really stuck with either -2, or 0 when it is an element.

So, what are the oxidation states of the Nitrogen on both sides of the equation? Has it gained or lost electrons (OIL,RIG)?

Which other part of the equation has lost/gained electrons?

Reply 19

OP