# Balancing redox reaction with one reactant

Consider the following acidic reaction:

$\ce{As -> H2AsO4- + AsH3}$

How would I go about balancing this using the half reaction/ion-electron method?

• As is the reactant for both half reactions, yielding $\ce{As->H2AsO4^-}$ and $\ce{As->AsH3}$ – bobthechemist Sep 23 '13 at 1:07

As bobthechemist mentioned in his comment, use the reactant species in both half reactions.

Consider this example, in which the same nitrogen species $\ce{N2}$ is used as the product in both half reactions:

$$\ce{NH4NO3-> N2}$$

Half reactions:

$$\ce{NH4+ ->N2}$$ $$\ce{NO3- ->N2}$$

Balancing the reduction half reaction: $$\ce{2NO3- ->N2}$$ $$\ce{2NO3- ->N2}\color{red}{\ce{+6H2O}}$$ $$\ce{2NO3-}\color{red}{\ce{+12H+}}\ce{ ->N2 +6H2O}$$ $$\ce{2NO3- +12H+}\color{red}{\ce{+10e-}}\ce{ ->N2 +6H2O}$$

Balancing the oxidation half reaction: $$\ce{2NH4+ ->N2}$$ $$\ce{2NH4+ ->N2}\color{red}{\ce{+8H+}}$$ $$\ce{2NH4+ ->N2 +8H+}\color{red}{\ce{+8e-}}$$

Combining the equations: $$\color{red}{4}({\ce{2NO3- +12H+ +10e- ->N2 + 6H2O}})$$ $$\underline{\color{red}{+5}(\ce{2NH4+ ->N2 +8H+ +8e-})}$$ $$\ce{8NO3- + 10NH4+ +48H+ + 40e- + 10H2O -> 9N2 +24H2O +40H+ + 40e- }$$ $$\ce{8NO3- + 10NH4+ +8H+ -> 9N2 +14H2O}$$