# What are the coefficients in front of NO3-(aq) and Cu(s) when the following redox equation is balanced in an acidic solution: ___ NO3-(aq) + ___ Cu(s) → ___ NO(g) + ___ Cu2+(aq)? A) 2, 6 B) 3, 6 C) 3, 4 D) 2, 3

Apr 16, 2015

Your reaction has dilute nitric acid reacting with copper metal to produce nitric oxide and $C {u}^{2 +}$ ions. The reaction will take place in acidic solution, which means you can add water and ${H}^{+}$ to balance oxygen and hydrogen.

So,

$\stackrel{\textcolor{b l u e}{+ 5}}{N}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{3 \left(a q\right)}^{-}}$ $+ \stackrel{\textcolor{b l u e}{0}}{C {u}_{\left(s\right)}}$ $\to \stackrel{\textcolor{b l u e}{+ 2}}{N}$ stackrel(color(blue)(-2))(O_((g)) $+ C {u}_{\left(a q\right)}^{2 +}$

Notice that nitrogen is reduced from a +5 to a +2 oxidation state, while copper is oxidized from a 0 to a +2 oxidation state.

$\stackrel{\textcolor{b l u e}{0}}{C u} \to \stackrel{\textcolor{b u e}{+ 2}}{C u} + 2 {e}^{-}$ $| \cdot 3$ $\to$ oxidation

$\stackrel{\textcolor{b l u e}{+ 5}}{N} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 2}}{N}$ $| \cdot 2$ $\to$ reduction

The balanced half-reactions will be

$3 \stackrel{\textcolor{b l u e}{0}}{C u} \to 3 \stackrel{\textcolor{b u e}{+ 2}}{C u} + 6 {e}^{-}$

$2 \stackrel{\textcolor{b l u e}{+ 5}}{N} + 6 {e}^{-} \to 2 \stackrel{\textcolor{b l u e}{+ 2}}{N}$

Before balancing the oxygen and hydrogen atoms ,the equation will look like this

$2 \stackrel{\textcolor{b l u e}{+ 5}}{N}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{3 \left(a q\right)}^{-}}$ $+ 3 \stackrel{\textcolor{b l u e}{0}}{C {u}_{\left(s\right)}}$ $\to 2 \stackrel{\textcolor{b l u e}{2}}{N}$ stackrel(color(blue)(-2))(O_((g)) $+ 3 C {u}_{\left(a q\right)}^{2 +}$

SIDE NOTE Here's how you'd go about balancing the chemical equation. First, balance the oxygen atoms. Notice you have 6 oxygen atoms on the ractants' side and only 2 on the products' side $\to$ add 4 water molecules on the products' side.

$2 \stackrel{\textcolor{b l u e}{+ 5}}{N}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{3 \left(a q\right)}^{-}}$ $+ 3 \stackrel{\textcolor{b l u e}{0}}{C {u}_{\left(s\right)}}$ $\to 2 \stackrel{\textcolor{b l u e}{2}}{N}$ stackrel(color(blue)(-2))(O_((g)) $+ 3 C {u}_{\left(a q\right)}^{2 +} + 4 {H}_{2} {O}_{\left(l\right)}$

FInally, balance the hydrogen atoms by adding 8 ${H}^{+}$ on the reactants' side

$8 {H}_{\left(a q\right)}^{+} + 2 \stackrel{\textcolor{b l u e}{+ 5}}{N}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{3 \left(a q\right)}^{-}}$ $+ 3 \stackrel{\textcolor{b l u e}{0}}{C {u}_{\left(s\right)}}$ $\to 2 \stackrel{\textcolor{b l u e}{2}}{N}$ stackrel(color(blue)(-2))(O_((g)) $+ 3 C {u}_{\left(a q\right)}^{2 +} + 4 {H}_{2} {O}_{\left(l\right)}$