Question

When the following equation is balanced properly under acidic conditions, what are the coefficients of the species shown?CIO3 +NO 2-CI +Water appears in the balanced equation as aNO3How many electrons are transferred in this reaction?(reactant, product, neither) with a coefficient of(Enter 0 for neither.)

Answer 1

Answer:

Explanation:

Here, we want to balance the given equation in acidic conditions

We start by breaking the equation into half equations

We have that as:

$\begin{gathered} ClO_3^-\text{ }\rightarrow\text{ Cl}^- \\ NO_2\rightarrow\text{ NO}_3^- \end{gathered}$

The next step is to balance other elements aside from oxygen and hydrogen

This is already done

The next step is to balance oxygen by adding H2O:

$\begin{gathered} ClO_3^-\text{ }\rightarrow\text{ Cl}^-\text{ +3 H}_2O \\ NO_2\text{ + H}_2O\text{ }\rightarrow NO_3^-\text{ } \end{gathered}$

The next step is to balance hydrogen by adding the hydrogen ion

We have that as:

$\begin{gathered} ClO_3^-\text{ + 6H}^+\text{ }\rightarrow Cl^-\text{ + 3H}_2O \\ NO_2\text{ + H}_2O\text{ }\rightarrow\text{ NO}_3^-\text{ + 2H}^+ \end{gathered}$

The next step here is to balance the excess ions by adding electrons:

$\begin{gathered} ClO_3^-\text{ + 6H}^+\text{ + 4e}^-\rightarrow\text{ Cl}^-\text{ + 3H}_2O \\ NO_2\text{ + H}_2O\text{ }\rightarrow\text{ NO}_3^-\text{ + 2H}^+\text{ + e}^- \end{gathered}$

Now, we need to cancel out the electrons.

We multiply equation ii by 4:

$\begin{gathered} ClO_{3\text{ }}^-\text{ + 6H}^+\text{ + 4e}^-\text{ }\rightarrow\text{ Cl}^-\text{ + 3H}_2O \\ 4NO_2\text{ + 4H}_2O\text{ }\rightarrow\text{ 4NO}_3^-\text{ + 8H}^+\text{ + 4e}^- \end{gathered}$

Now, let us merge the two equations:

$ClO_3^-\text{ + 4NO}_2\text{ + H}_2O\text{ }\rightarrow\text{ Cl}^-\text{ + 4NO}_3^-\text{ + 2H}^+$