Chemistry Forum

For example, why in reaction FeS₂ + O₂ --> Fe₂O₃ + SO₂ we double the sulfur electrons from 5 to 10 but don't double the electron of the iron despite the fact we have two iron atoms on the right?
Or, e. g., the reaction K₄Fe(CN)₆ + KMnO₄ + H₂SO₄ --> KHSO₄ + Fe₂(SO₄)₃ + MnSO₄ + HNO₃ + CO₂ + H₂O. Should we double the electron of the iron or to leave it as is?

I dont understand exactly what are you asking... Are you asking for general rules how to balance equations or is it something else? eg- "Double sulfur electrons?" what do you mean by this?
Sry english is not my native language.

Alex C. wrote:For example, why in reaction FeS₂ + O₂ --> Fe₂O₃ + SO₂ we double the sulfur electrons from 5 to 10 but don't double the electron of the iron despite the fact we have two iron atoms on the right?

I have no idea what you are talking about.

Each sulfur atom loses five electrons, each iron atom losses one electron, and each oxygen atom gains two.

Since the Persulfide and Ferrous ions must be the same amount and are the only source of iron and sulfur, we have only two unknowns: the amount of iron (x) and the amount of oxygen (y).

The RedOx half equations are:

Oxidation:
x Fe²⁺ = x Fe³⁺ + x {e-}
x S2^2- = 2x S⁴⁺ + 10x {e-}
Reduction:
y O2 + 4y{e-} = 2y O{2-)

Balancing the electrons gives the lowest integer solution of: x=4, y=11

Substututing back in and combining (and distributing oxygen appropriately):

4 FeS2 + 11 O2 = 2 Fe2O3 + 8 SO2

Alex C. wrote:Or, e. g., the reaction K₄Fe(CN)₆ + KMnO₄ + H₂SO₄ --> KHSO₄ + Fe₂(SO₄)₃ + MnSO₄ + HNO₃ + CO₂ + H₂O. Should we double the electron of the iron or to leave it as is?

If you're asking is the iron oxidized from ferrous to ferric (2+ to 3+), then yes it is. This does not double the electron; you remove one from each atom.

If you’re asking how to balance the monster equation, then you could break it into half equations. However, it’s easier to just balance atoms in this case.

Assign variables to Fe, C, N, Mn, K and H (simplified to a minimum number of unknowns.)

Let Fe be 2z (since there is Fe₂).
2z K4Fe(CN)6 + ? KMnO4 + ? H2SO4 --> ? KHSO4 + z Fe2(SO4)3 + ? MnSO4 + ? HNO3 + ? CO2 + ? H2O

This enumerates C and N.
2z K4Fe(CN)6 + ? KMnO4 + ? H2SO4 --> ? KHSO4 + z Fe2(SO4)3 + ? MnSO4 + 12z HNO3 + 12z CO2 + ? H2O

Let Mn be 2y (experience)
2z K4Fe(CN)6 + 2y KMnO4 + ? H2SO4 --> ? KHSO4 + z Fe2(SO4)3 + 2y MnSO4 + 12z HNO3 + 12z CO2 + ? H2O

From LHS we have K := 8z + 2y.
2z K4Fe(CN)6 + 2y KMnO4 + ? H2SO4 --> (8z+2y) KHSO4 + z Fe2(SO4)3 + 2y MnSO4 + 12z HNO3 + 12z CO2 + ? H2O

Let H2O be x, giving total H of (20z+2y+2x)
2z K4Fe(CN)6 + 2y KMnO4 + (10z+y+x) H2SO4 --> (8z+2y) KHSO4 + z Fe2(SO4)3 + 2y MnSO4 + 12z HNO3 + 12z CO2 + x H2O

Balance for SO4: x = z+3y

Balance for other O: 8y = 60z +x

Substitute and simplify: 5y=61z

Thus the Minimum Integer Solution: z=5, y=61, x=188

10 K4Fe(CN)6 + 122 KMnO4 + 299 H2SO4 --> 162 KHSO4 + 5 Fe2(SO4)3 + 122 MnSO4 + 60 HNO3 + 60 CO2 + 188 H2O

This is a classic exam question, by the way.