[joshuaspence]

Ok... I am having trouble balancing the following reaction:

            2MnO2(s) + H2(g) ? Mn2O3(s) + H2O(l)

I know if you add up the charges you get a net charge of 0 on both sides (at least I think so)

BUT

If you add up the number of electrons on each side, there are 2 more electrons on the right hand side than the left hand side. I have pondered for hours... obviously the electrons cannot be "created", they must come from somewhere....

Where have I gone wrong?

Electrons on Left hand side:
2Mn4+ = 42 electrons
2(O2)2- = 36 electrons
H2 = 2 electrons
Total on LHS = 80 electrons

Electrons on right hand side:
(Mn3+)2 = 44 electrons
(O2-)3 = 24 electrons
(H+)2 = 0 electrons
O2- = 8 electrons
Total on RHS = 76 electrons

[tbuihuu]

I think you wrong.
Electrons on Left hand side:
2Mn4+ =2x(25-4)= 42 electrons
2(O2-)2 =2x2x(8+2=) 40 electrons
H2 = 2 electrons
Total on LHS = 84 electrons

Electrons on right hand side:
(Mn3+)2 = 2x(25-3)=44 electrons
(O2-)3 = 3x(8+2)=30 electrons
(H+)2 =2x(1-1)= 0 electrons
O2- = 8+2=10 electrons
Total on RHS = 84 electrons

[Borek]

I know if you add up the charges you get a net charge of 0 on both sides (at least I think so)

Not necesarilly. You must have the same charge on both sides, but it doesn't have to be zero.

Zn + Cu²⁺ -> Zn²⁺ + Cu

+2 on both sides.

When calculating charges don't check all electrons - look just for charges of the molecules/ions present. So in the example above don't count zinc as 30, but as 0 (on the left) or +2 (on the right). All other electrons cancel out.

[joshuaspence]

I think you wrong.
Electrons on Left hand side:
2Mn4+ =2x(25-4)= 42 electrons
2(O2-)2 =2x2x(8+2=) 40 electrons
H2 = 2 electrons
Total on LHS = 84 electrons

Electrons on right hand side:
(Mn3+)2 = 2x(25-3)=44 electrons
(O2-)3 = 3x(8+2)=30 electrons
(H+)2 =2x(1-1)= 0 electrons
O2- = 8+2=10 electrons
Total on RHS = 84 electrons

Yep you are right on this one....

But something else I cant understand...

If you add up the valence electrons only


Electrons on Left hand side:
2Mn = 2 * 4 = 8 electrons
2O2 = 2*2*6 = 24 electrons
H2 = 1*2 = 2 electrons
Total on LHS = 34 electrons

Electrons on right hand side:
Mn2 = 2*5 = 10 electrons
O3 = 3*6 =18 electrons
H2 = 2 electrons
O = 6 electrons
Total on RHS = 36 electrons

 

Thanks in advance

[Borek]

If you add up the valence electrons only

Electrons on Left hand side:
2Mn = 2 * 4 = 8 electrons

Electrons on right hand side:
Mn2 = 2*5 = 10 electrons

Please elaborate. Note: both oxides are neutral, so they contain exactly the same number of electrons (both total and valence) as sum of separately treated neutral atoms of elements they are composed of.

[joshuaspence]

If you add up the valence electrons only

Electrons on Left hand side:
2Mn = 2 * 4 = 8 electrons

Electrons on right hand side:
Mn2 = 2*5 = 10 electrons

Please elaborate. Note: both oxides are neutral, so they contain exactly the same number of electrons (both total and valence) as sum of separately treated neutral atoms of elements they are composed of.

The manganese has been reduced, and hence gained electrons....

The oxides are different: one is Manganese (IV) oxide, the other is dimanganese (III) trioxide

[Borek]

The manganese has been reduced, and hence gained electrons....

Hydrogen has been oxidised, and hence lost electrons That explains where you have lost 2 electrons.

Note: if you take several neutral atoms and they react creating a new neutral molecule, number of electrons present on the molecular orbitals is exactly the same as number of valence electrons in isolated atoms. Thus when calculating electrons in th emolecule it doesn't matter what the atom valence is. Sometimes - especially in the case of ionic compounds - you may assign individual electrons to individual atmos (ions) - but even then when you look at the smallest possible neutral ionic "molecule" (like NaCl) it has the same number of valence electrons as isolated atoms had before - Cl gained one electron, Na lost one, total number of electrons didn't change.

[joshuaspence]

The manganese has been reduced, and hence gained electrons....

Hydrogen has been oxidised, and hence lost electrons That explains where you have lost 2 electrons.

Note: if you take several neutral atoms and they react creating a new neutral molecule, number of electrons present on the molecular orbitals is exactly the same as number of valence electrons in isolated atoms. Thus when calculating electrons in th emolecule it doesn't matter what the atom valence is. Sometimes - especially in the case of ionic compounds - you may assign individual electrons to individual atmos (ions) - but even then when you look at the smallest possible neutral ionic "molecule" (like NaCl) it has the same number of valence electrons as isolated atoms had before - Cl gained one electron, Na lost one, total number of electrons didn't change.

Im not quite sure I understand that....

It just seems logical that the sum of the of electrons in the outermost shells should remain constant on both sides of the equation... Unless an electron was excited to a higher orbital or vice versa?

This thought arose when trying to draw Lewis diagrams for this reaction.

I have attached my Lewis diagrams... The top is the reactants and the  bottom is the products. The gap inbetweeen the Manganese atoms in the products is where there should be 2 extra electrons, for two reason:

1. To complete the octet of the manganese atoms
2. To correctly label the manganese atoms as manganese (III) not manganese (IV)

[tbuihuu]

hey,  the number valance electrons of Mn is 7 (3d⁵4s²)

and
 formula of  Mn₂O₃ is O=Mn-O-Mn=O
you can see again but i think in reaction all thing is balance
good luck!

[joshuaspence]

hey,  the number valance electrons of Mn is 7 (3d⁵4s²)

and
 formula of  Mn₂O₃ is O=Mn-O-Mn=O
you can see again but i think in reaction all thing is balance
good luck!

correct me if im wrong... but valence electrons is the outermost shell only... not incomplete subshells of inner shells

and ok well iv redrawn the diagram... but it still is missing 2 electrons (as indicated by lowercase e's)

what have i done wrong?

[joshuaspence]

actually i just noticed my diagram is wrong... i cant draw the lewis diagram for O=Mn-O-Mn=O

 

anyone know how?

[Borek]

correct me if im wrong... but valence electrons is the outermost shell only... not incomplete subshells of inner shells

Not in the case of transition elements.

Part of your problem is that you are trying to use terms and methods (like definition of valence electrons and Levis diagrams) outside of their range of application. Transition elements use their d electrons when combining with other elements (check possible valences of Mn or Cr - they can't be explained with s & p electrons alone), when d electrons come into play, octet rule becomes 18-electron rule and Levis diagrams are of no use.

[joshuaspence]

Thanks for your help...

This is all too confusing for a year 11 chemistry student... hehe