# Balancing redox equation with oxygen on both sides?

The problem is to balance this equation in an acidic medium.

$$\ce{As2O3 (s) + NO3- (aq) -> H3AsO4 (aq) + N2O3 (aq) }$$

However, I am getting stuck balancing the oxygens after dividing them into half reactions.

How do you balance a redox reaction when oxygen is involved in both sides of the equation?

• How do you balance it when oxygen (or any element, for that matter) is not present on either side? You simply won't be able to balance it then. That's the very idea of balancing. You must have as much oxygen (and any other element too) on the left as on the right. – Ivan Neretin Apr 30 at 15:21
• Refer the following link. periodni.com/… – Ak19 Apr 30 at 15:22

It is always easy to balance redox equations by considering two half reactions. The two relevant half reactions to your redox equation are: \begin{align} \ce{As2O3 (s) -> H3AsO4 (aq)} \\ \ce{NO3- (aq) -> N2O3 (aq)}\\ \end{align}
First balance $$\ce{As}$$ and $$\ce{N}$$ in both equations. Then, balance $$\ce{O}$$ in both equations with water and $$\ce{H}$$ with $$\ce{H+}$$:
\begin{align} \ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq)}\\ \ce{2 NO3- (aq) + 6 H+ (aq) -> N2O3 (aq) + 3 H2O (l)}\\ \end{align}
Finally, balance the charges by electrons: \begin{align} \ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq) + 4 e-}\\ \ce{2 NO3- (aq) + 6 H+ (aq) + 4 e- -> N2O3 (aq) + 3 H2O (l)}\\ \end{align}
Now add the last mass and charge balanced two equations in order to cancel electrons (this is easy, because both have 4 $$\ce{e-}$$s):
$$\ce{As2O3 (s) + 2 NO3- (aq) + 2 H+ (aq) + 2 H2O (l) -> 2 H3AsO4 (aq) + N2O3 (aq)}$$ Or simply, $$\ce{As2O3 (s) + 2 NO3- (aq) + 2 H3O+ (aq) -> 2 H3AsO4 (aq) + N2O3 (aq)}$$