Use the stated reaction medium and products when balancing a half-equation. In neutral aqueous conditions, oxygen reduction is commonly written O2+2H2O+4e−→4OH−. This produces hydroxide locally; it does not require an initially concentrated alkaline solution. Hydrogen formation from water can be written 2H2O+2e−→H2+2OH−.
For a specified neutral or mildly alkaline permanganate reduction to MnO2, the balanced form is MnO4−+2H2O+3e−→MnO2+4OH−. In explicitly alkaline equations OH− can be a reactant, as in 4OH−→O2+2H2O+4e−. Check atoms and charge separately. Electrons belong on the reactant side for reduction and product side for oxidation.
Example 1
Balance O2 reduction to OH− in neutral water.
O2+2H2O+4e−→4OH−. Each side has four O and four H atoms, with charge −4.
Example 2
Balance the specified MnO4−→MnO2 reduction in neutral water.
MnO4−+2H2O+3e−→MnO2+4OH−. Each side contains Mn1O6H4 and has total charge −4.
Watch out:The medium and specified products matter; neutral or alkaline half-equations can use water and hydroxide.
Half-equations in neutral and alkaline water practice | Year 11 Chemistry | Ascent Arcade