We will investigate the rules for balancing redox reactions by applying these

rules to the following chemical reaction Fe2+ + Cr2O72- ---> Fe3+ + Cr3+, in both

1. Determine Substance Oxidized and Substance Reduced and Write in

First, determine which substance is being oxidized and which is being reduced.

Separate them into two separate half-reactions: one for oxidation and one for

reduction. In our example, Fe is being oxidized, (going from +2 to +3), and Cr is

being reduced, (going from +6 in Cr2O72- to +3 in Cr+3). Our half-reactions are thus

Fe2+ ---> Fe3+ oxidation half-reaction

Cr2O72- ---> Cr3+ reduction half-reaction

2. Balance All Elements Except O and H:

The first reaction is fine as is, but the second needs to have the Cr atoms

balanced. After this step our two reactions are as follows:

Fe2+ ---> Fe3+ oxidation half-reaction

Cr2O72- ---> 2Cr3+

Fe2+ ---> Fe3+ oxidation half-reaction

H+ ions to the other side to balance the H atoms. We will need to add 14 H+

6Fe2+ + Cr2O72- + 14H20 ---> 6Fe3+ 2Cr3+ + 7 H20 + 14 OH-

in the oxidation half-reaction:

2. Caution: The Fe atom is neither oxidized nor reduced. It is +2 on both sides.

products. I.e., however many electrons are needed to reduce the oxidizing

the same number of electrons.

Some topics in this essay:
Charge Half-Reaction, Balance Atoms, Electrons Cancel, Separate Half-Reactions, OK H+, Practice Try, Basic Solution, Acidic Solution, Reactants Products, Finally H2O, oxidation half-reaction, reduction half-reaction, 2cr3+ +, 2cr3+ + 7, + 7 h20, 7 h20, + 7, cr2o72- +, oxidation half-reaction cr2o72-, ---> fe3+, h+ ions, half-reaction cr2o72-, + 4h2o, cr2o72- + 14h+, fe2+ ---> fe3+,