Question

cr2o72-+fe2+=cr3++fe3+
balance by oxidation no. method​

Answer 1

Answer:

Cr2O72- + Fe2+ → Cr3+ + Fe3+

1. Split up into two half reactions for each of the elements (ignore hydrogen or oxygen, unless they’re free-standing elements and you have no other choice)

For this one, that means you take the chromium and iron and separate them

Cr2O72-→ Cr3+

Fe2+ → Fe3+

2. Balance the number of all atoms besides hydrogen and oxygen

Cr2O72-→ 2Cr3+

We need a coefficient 2 on the product side because Cr2O72- has an extra chromium

Fe2+ → Fe3+

This one is fine since there is 1 iron on each side

3. Balance all the oxygens by adding an H2O for each extra oxygen you need

Cr2O72-→ 2Cr3+ + 7H2O

Since Cr2O72- has 7 oxygens, we add 7 water molecules to the products to balance it out

Fe2+ → Fe3+

This one is fine since there are no oxygen atoms

4. Balance all the hydrogens by adding +14 H+ ions for each extra hydrogen you need

Cr2O72-+ 14H+ → 2Cr3+ + 7H2O

Because of the 7 water molecules we added, we need 14 hydrogen ions to balance out the reactants

Fe2+ → Fe3+

This one is fine since there are no hydrogen atoms

5. Balance the charges by adding electrons to the more positive side

Cr2O72-+ 14H+ + 6e- → 2Cr3+ + 7H2O

This one’s a little tricky…

The overall charge of the reactants is +12, because of the 14H+ and the one Cr2O72-

The overall charge of the products is +6 because of the 2Cr3+

Therefore, we add 6 electrons to the reactants so that both sides are +6

Fe2+ → Fe3+ + e-

The reactants are +2 and the products are +3, so the products need one electron

Notice how this reaction has electrons as products and the first one has electrons as reactants

6. Make it so that each of the two half reactions has the same number of electrons (kind of like systems of equations in algebra)

Cr2O72-+ 14H+ + 6e- → 2Cr3+ + 7H2O

This one has more, so just leave it the same

6 (Fe2+ → Fe3+ + e-) → 6Fe2+ → 6Fe3+ + 6e-

we multiply everything by 6

7. Combine the half-reactions and simplify whatever you can

6Fe2+ Cr2O72-+ 14H+ + 6e- → 2Cr3+ + 7H2O + 6Fe3+ + 6e-

6Fe2+ Cr2O72-+ 14H+ → 2Cr3+ + 7H2O + 6Fe3+

The electrons should always cancel out for your final answer