[Pede]

I need to balance the following:

CH3CH2OH + Cr2O7{-2} + H{+} = CH3CHO + Cr{3+} + H2O

Oxidation numbers
C oxidated by 2
Cr reduced by 3

3 CH3CH2OH + Cr2O7{-2} + H{+} = 3 CH3CHO + 2 Cr{3+} + H2O

I need to balance the remaining H and water, any help? been using half an hour on this now .

[sjb]

I need to balance the following:

CH3CH2OH + Cr2O7{-2} + H{+} = CH3CHO + Cr{3+} + H2O

Oxidation numbers
C oxidated by 2
Cr reduced by 3

3 CH3CH2OH + Cr2O7{-2} + H{+} = 3 CH3CHO + 2 Cr{3+} + H2O

I need to balance the remaining H and water, any help? been using half an hour on this now .

Firstly, no need to crosspost.

Secondly, it may be easier if you try and balance the half equations individually

Cr₂O₇^2-  Cr³⁺; and
MeCH₂OH  MeCHO

[Pede]

Secondly, it may be easier if you try and balance the half equations individually

Cr₂O₇^2-  Cr³⁺; and
MeCH₂OH  MeCHO

Sorry, but isn't that what I've already done? I just need to find out how much H+ and water that I need to add to both sides.

[sjb]

Sorry, but isn't that what I've already done? I just need to find out how much H+ and water that I need to add to both sides.

Well, you have, sort of, yes.

I find it easier to do things like

MeCH₂OH  MeCHO + 2H⁺ + 2e^-, and similar for the dichromate species, then balance the electrons.

So for Cr₂O₇^2- Cr³⁺, first balance the chromium, then the oxygen and protons, and add electrons to balance the charge. Then see how the electrons can cancel out by lowest common multiples of each equation.

[Pede]

Sorry, but isn't that what I've already done? I just need to find out how much H+ and water that I need to add to both sides.

Well, you have, sort of, yes.

I find it easier to do things like

MeCH₂OH   MeCHO + 2H⁺ + 2e^-, and similar for the dichromate species, then balance the electrons.

So for Cr₂O₇^2- Cr³⁺, first balance the chromium, then the oxygen and protons, and add electrons to balance the charge. Then see how the electrons can cancel out by lowest common multiples of each equation.

That sounds smart, but never done it like that before, I am not sure that I get it, but I'll try:


3 CH3CH2OH + Cr2O7{-2} + H{+} = 3 CH3CHO + 2 Cr{3+} + H2O

3 CH3CH2OH   3 CH3CHO + 6 H⁺ + 6e^-

3 CH3CH2OH + Cr2O7{-2} + 6 H{+} 3 CH3CHO + 2 Cr{3+} + 6 H2O
Can anyone confirm if that is correct? :O


EDIT:
Figured it out, it's
3 CH3CH2OH + Cr2O7{-2} + 8 H{+} 3 CH3CHO + 2 Cr{3+} + 14 H2O

Got help from someone else. No need for help in this topic anymore.

[AWK]

3CH₃CH₂OH + Cr₂O₇^2- + 6H⁺ = 3CH₃CHO + 2 Cr³⁺ + 6H₂O
Can anyone confirm if that is correct? :O

You can simply check your solution - count sum of charges and all kind of atoms on both sides of equation - charges and oxygen atoms are incorect at the moment - the only way you can correct them is addition more H⁺ on the left side and water on the right side.

[Evaldas]

How, for example, H ion, without having reacted with something could have a charge? It should be H⁰, not H⁺, or not?

[AWK]

How, for example, H ion, without having reacted with something could have a charge? It should be H⁰, not H⁺, or not?

This is a reaction in acidic medium. You may use H₂SO₄ ans K₂Cr₂O₇ or Cr₂O₇^2- and H⁺