# Question d49b8

May 2, 2017

Warning! Long Answer. The coefficients are 2,3.

#### Explanation:

Here's how you get them.

The skeleton equation is

$\text{NO"_3^"-" + "Zn" → "Zn"^"2+" + "NO}$

Step 1. Separate the equation into two half-reactions.

$\text{NO"_3^"-" → "NO}$
$\text{Zn" → "Zn"^"2+}$

Step 2. Balance all atoms other than $\text{H}$ and $\text{O}$.

Done.

Step 3. Balance $\text{O}$ by adding $\text{H"_2"O}$ molecules to the deficient side.

$\text{NO"_3^"-" → "NO" +2"H"_2"O}$
$\text{Zn" → "Zn"^"2+}$

Step 4. Balance $\text{H}$ by adding $\text{H"^"+}$ ions to the deficient side.

$\text{NO"_3^"-" + "4H"^"+" → "NO" +2"H"_2"O}$
$\text{Zn" → "Zn"^"2+}$

Step 5. Balance charge by adding electrons to the deficient side.

$\text{NO"_3^"-" + "4H"^"+" +"3e"^"-" → "NO" +2"H"_2"O}$
$\text{Zn" → "Zn"^"2+" + "2e"^"-}$

Step 6. Multiply each half-reaction by a number to equalize the electrons transferred.

2×["NO"_3^"-" + "4H"^"+" +"3e"^"-" → "NO" +2"H"_2"O"]
3×["Zn" → "Zn"^"2+" + "2e"^"-"]

Step 7. Add the two half-reactions.

$2 \text{NO"_3^"-" + "3Zn" + "8H"^"+" → "2NO" + "3Zn"^"2+" + "4H"_2"O}$

Step 8. Convert to basic solution my adding the appropriate multiples of $\text{H"^"+" + "OH"^"-" → "H"_2"O}$ or $\text{H"_2"O" → "H"^"+" + "OH"^"-}$.

We want to cancel $\text{8H"^"+}$, so we add eight of the second equation.

2"NO"_3^"-" + "3Zn" + color(red)(cancel(color(black)("8H"^"+"))) → "2NO" + "3Zn"^"2+" + color(red)(cancel(color(black)("4H"_2"O")))
$\stackrel{\textcolor{b l u e}{4}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}} \text{H"_2"O" → color(red)(cancel(color(black)("8H"^"+"))) + "8OH"^"-}$
stackrel(————————————————————)(2"NO"_3^"-" + "3Zn" + 4"H"_2"O" → "2NO" + "3Zn"^"2+" + "8OH"^-")#

Step 9. Check that all atoms are balanced.

$\boldsymbol{\text{Atom"color(white)(m)"On the left"color(white)(m)"On the right}}$
$\textcolor{w h i t e}{m} \text{N} \textcolor{w h i t e}{m m m m m} 2 \textcolor{w h i t e}{m m m m m m m} 2$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m l} 10 \textcolor{w h i t e}{m m m m m m l} 10$
$\textcolor{w h i t e}{m} \text{Zn} \textcolor{w h i t e}{m m m m l} 3 \textcolor{w h i t e}{m m m m m m m} 3$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m l l} 8 \textcolor{w h i t e}{m m m m m m m} 8$

Step 10. Check that charge is balanced

$\boldsymbol{\text{On the left"color(white)(m)"On the right}}$
$\textcolor{w h i t e}{m m m} \text{2-"color(white)(mmmmmm)"2-}$

Everything checks! The equation is balanced.