# Balance the following redox reaction in acidic solution?

## $\setminus \textsf{{P}_{4} \left(s\right) + N {O}_{3}^{-} \setminus \to {H}_{2} P {O}_{4}^{-} \left(a q\right) + N O \left(g\right)}$ I have the oxidation, but not the reduction. Answer key tells me there is $\setminus \texttt{2 {H}_{2} O}$ but that doesn't match with the $\setminus \texttt{N {O}_{3}^{-}}$ on the other end of the arror. Plus, why add $\setminus \texttt{3 {e}^{-}}$? Bounded oxygen typically has a charge of -2, right? So three oxygens equals six electrons...?

Aug 9, 2018

And so elemental phosphorus $P \left(0\right)$ is oxidized to phosphoric acid $P \left(V\right)$, a five electron oxidation....

#### Explanation:

$\frac{1}{4} {P}_{4} \left(s\right) + 4 {H}_{2} O \rightarrow H P {O}_{4}^{2 -} + 7 {H}^{+} + 5 {e}^{-}$ $\left(i\right)$

Charge and mass are balanced, so this is kosher.

Nitrate $N \left(+ V\right)$ is REDUCED to $N O$ $N \left(+ I I\right)$.. . (Oxygen is slightly more electronegative than nitrogen, so it FORMALLY gets the electrons of the bonds. What is the oxidation number of oxygen in the REAL molecule, $O {F}_{2}$?)

$N {O}_{3}^{-} + + 4 {H}^{+} + 3 {e}^{-} \rightarrow N O + 2 {H}_{2} O$ $\left(i i\right)$

We add $3 \times \left(i\right) + 5 \times \left(i i\right)$ to get..

$\frac{3}{4} {P}_{4} \left(s\right) + 5 N {O}_{3}^{-} + 20 {H}^{+} + 15 {e}^{-} + 12 {H}_{2} O \rightarrow 3 H P {O}_{4}^{2 -} + 21 {H}^{+} + 15 {e}^{-} + 5 N O + 10 {H}_{2} O$

...and cancel away...

$\frac{3}{4} {P}_{4} \left(s\right) + 5 N {O}_{3}^{-} + 2 {H}_{2} O \rightarrow 3 H P {O}_{4}^{2 -} + 5 N O + 1 {H}^{+}$

Is this balanced? Should it be? I ask because anybody can butcher the arifmetik. We know that if it is NOT balanced with respect to mass and charge, it cannot be accepted as representation of reality.

Oxygen is not involved in redox here...phosphorus is, and nitrogen ONLY.... And all I have really done is to assign oxidation states.

Aug 9, 2018

Final balanced equation:
$3 {P}_{4} \left(s\right) + 20 N {O}_{3}^{-} \left(a q\right) + 8 {H}^{+} \left(a q\right) + 8 {H}_{2} O \left(l\right) \to 20 N O \left(g\right) + 12 {H}_{2} P {O}_{4}^{-} \left(a q\right)$

#### Explanation:

Here we go:

In an acidic solution means I can strategically place hydrogen ion and water wherever in the equation necessary:

P_4(s)+16H_2O->4H_2PO_4^-)+20e^- )+24H^+

NO_3^-)+4H^(+)+ 3e^(- )->NO+2H_2O

Why add $3 {e}^{-}$?
Because ${N}^{+ 5}$ is reduced to ${N}^{+ 2}$

The electrons on each side must be equal:

$3 \left({P}_{4} \left(s\right) + 16 {H}_{2} O \to 4 {H}_{2} P {O}_{4}^{-} + 20 {e}^{-} + 24 {H}^{+}\right)$
20(NO_3^-)+4H^(+)+ 3e^(- )->NO+2H_2O)

3P_4(s)+48H_2O->12H_2PO_4^-)+60e^- )+72H^+
$20 N {O}_{3}^{-} + 80 {H}^{+} + 60 {e}^{-} \to 20 N O + 40 {H}_{2} O$

I am left with:
$3 {P}_{4} \left(s\right) + 20 N {O}_{3}^{-} \left(a q\right) + 8 {H}^{+} \left(a q\right) + 8 {H}_{2} O \left(l\right) \to 20 N O \left(g\right) + 12 {H}_{2} P {O}_{4}^{-} \left(a q\right)$