When balancing half equations of redox reactions, $\ce{H2O}$ is used to balance any oxygen atoms? For example, this half equation:

$$\ce{NO3- → NH4+}$$

is balanced into:

$$\ce{NO3- + 8 e- + 10 H+ → NH4+ + 3 H2O}$$

Why is $\ce{OH-}$ not used instead of $\ce{H2O}$? E.g. The balanced equation becomes

$$\ce{NO3- + 8 e- + 7 H+ → NH4+ + 3 OH-}$$

When balancing half equations of redox reactions, $\ce{H2O}$ is used to balance any oxygen atoms? For example, this half equation

$$\ce{NO3- → NH4+}$$

is balanced into:

$$\ce{NO3- + 8 e- + 10 H+ → NH4+ + 3 H2O}$$

Why is $\ce{OH-}$ not used instead of $\ce{H2O}$? E.g. the balanced equation becomes

$$\ce{NO3- + 8 e- + 7 H+ → NH4+ + 3 OH-}$$

When balancing half equations of redox reactions, H₂O is used to balance any oxygen atoms. For example, this half equation:

NO₃^- -> NH₄⁺

is balanced into:

8e^- + NO₃^- + 10H⁺ -> NH₄⁺ + 3H₂O

Why is OH^- not used instead of H₂O?

(E.g. The balanced equation becomes

8e^- + NO₃^- + 7H⁺ -> NH₄⁺ + 3OH^-)

Thanks!

When balancing half equations of redox reactions, $\ce{H2O}$ is used to balance any oxygen atoms? For example, this half equation:

$$\ce{NO3- → NH4+}$$

is balanced into:

$$\ce{NO3- + 8 e- + 10 H+ → NH4+ + 3 H2O}$$

Why is $\ce{OH-}$ not used instead of $\ce{H2O}$? E.g. The balanced equation becomes

$$\ce{NO3- + 8 e- + 7 H+ → NH4+ + 3 OH-}$$