Copper reacts with dilute nitric acid to form copper(II) nitrate solution, nitrogen monoxide gas and water: Cu(s) + HNO3(aq) → Cu(NO3)2(aq) + NO(g) + H2O(l) (unbalanced) (a) State the oxidation number of copper in Cu(s), of nitrogen in HNO3(aq), and of nitrogen in NO(g). [2] (b)…

Answer: ((a)) Cu in Cu(s) is 0; N in HNO3(aq) is +5; N in NO(g) is +2 ((b)) Cu oxidised : N reduced = 3 : 2 (each Cu loses 2e^-, each reduced N gains 3e^-; lcm(2,3)=6 electrons transferred) ((c)) 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l); of the 8 N atoms, only 2 are reduced to NO, the other 6 remain as unchanged NO3^- spectator ions in 3Cu(NO3)2 (3×2=6) ((d)) Atoms: Cu 3=3; N 8=6+2=8; H 8=4×2=8; O 8×3=24=3×6+2×1+4×1=18+2+4=24. Charge: both sides are neutral molecules/formula units, so total charge is 0=0 on each side.. Part (a): Oxidation numbers Copper is an uncombined element, so its oxidation number is 0. In HNO3, hydrogen is +1 and oxygen is -2 (three oxygens give -6); for the neutral molecule, (+1) + x + (-6) = 0 ⇒ x = +5. So nitrogen in HNO3 is +5.…

AS Chemistry 9701: Balancing the Copper + Dilute Nitric Acid Equation by Oxidation Number

Copper reacts with dilute nitric acid to form copper(II) nitrate solution, nitrogen monoxide gas and water:

$\text{Cu}(s) + \text{HNO}_3(aq) \rightarrow \text{Cu(NO}_3)_2(aq) + \text{NO}(g) + \text{H}_2\text{O}(l) \quad \text{(unbalanced)}$

(a) State the oxidation number of copper in $\text{Cu}(s)$ , of nitrogen in $\text{HNO}_3(aq)$ , and of nitrogen in $\text{NO}(g)$ . [2]

(b) Using the change in oxidation number of copper and of the nitrogen atoms that are reduced, deduce the simplest whole-number ratio of moles of copper oxidised to moles of nitrogen atoms reduced, such that the total number of electrons lost equals the total number of electrons gained. [2]

(c) Hence deduce the balancing numbers for the full equation above, and explain how the number of $\text{HNO}_3(aq)$ formula units is obtained even though not all of the nitrogen atoms are reduced. [3]

(d) Verify that your balanced equation from (c) conserves both atoms and charge (noting that this is a reaction between neutral molecules, so the overall charge on each side should be zero). [2]

Part (a): Oxidation numbers

Copper is an uncombined element, so its oxidation number is $0$ .

In $\text{HNO}_3$ , hydrogen is $+1$ and oxygen is $-2$ (three oxygens give $-6$ ); for the neutral molecule, $(+1) + x + (-6) = 0 \implies x = +5$ . So nitrogen in $\text{HNO}_3$ is $+5$ .

In $\text{NO}$ , oxygen is $-2$ ; for the neutral molecule, $x + (-2) = 0 \implies x = +2$ . So nitrogen in $\text{NO}$ is $+2$ .

Part (b): Electron balance ratio

Copper: $0 \rightarrow +2$ , a rise of $2$ . Each Cu atom loses $2$ electrons.

Nitrogen (the atoms that end up as NO): $+5 \rightarrow +2$ , a fall of $3$ , each of these N atoms gains $3$ electrons.

For electrons lost to equal electrons gained, find the lowest common multiple of $2$ and $3$ , which is $6$ :

$3\ \text{Cu atoms} \times 2e^- = 6e^- \text{ lost} \qquad 2\ \text{N atoms} \times 3e^- = 6e^- \text{ gained}$

So the simplest whole-number ratio of copper oxidised to nitrogen reduced is $\boxed{3:2}$ .

Part (c): Balancing the full equation

Copper oxidised $:3$ , so place a $3$ in front of $\text{Cu}$ and, since each $\text{Cu}^{2+}$ needs two $\text{NO}_3^-$ ions, a $3$ in front of $\text{Cu(NO}_3)_2$ .

Nitrogen reduced $:2$ , so place a $2$ in front of $\text{NO}$ .

Counting nitrogen: the right-hand side now has $3 \times 2 = 6$ nitrogen atoms as spectator $\text{NO}_3^-$ ions (in $3\text{Cu(NO}_3)_2$ ) plus $2$ nitrogen atoms reduced to $\text{NO}$ , giving $6 + 2 = 8$ nitrogen atoms in total. Since every nitrogen atom on the right must have come from an $\text{HNO}_3$ molecule on the left, $8\text{HNO}_3$ is required. even though only $2$ of those $8$ nitrogen atoms are actually reduced; the other $6$ simply act as the acid, supplying $\text{H}^+$ and ending up unchanged as $\text{NO}_3^-$ ligands bound to $\text{Cu}^{2+}$ .

Balancing hydrogen and oxygen with water then gives $4\text{H}_2\text{O}$ :

$3\text{Cu}(s) + 8\text{HNO}_3(aq) \rightarrow 3\text{Cu(NO}_3)_2(aq) + 2\text{NO}(g) + 4\text{H}_2\text{O}(l)$

Part (d): Verifying atoms and charge

Atoms:

Cu: left $3$ ; right $3$ (in $3\text{Cu(NO}_3)_2$ ). $3 = 3$ ✓
N: left $8$ (in $8\text{HNO}_3$ ); right $3\times2 = 6$ (spectator, in $3\text{Cu(NO}_3)_2$ ) $+\ 2$ (reduced, in $2\text{NO}$ ) $= 8$ . $8 = 8$ ✓
H: left $8$ (in $8\text{HNO}_3$ ); right $4\times2 = 8$ (in $4\text{H}_2\text{O}$ ). $8 = 8$ ✓
O: left $8\times3 = 24$ (in $8\text{HNO}_3$ ); right $3\times6 = 18$ (in $3\text{Cu(NO}_3)_2$ , each with $6$ O) $+\ 2\times1 = 2$ (in $2\text{NO}$ ) $+\ 4\times1 = 4$ (in $4\text{H}_2\text{O}$ ) $= 18+2+4 = 24$ . $24 = 24$ ✓

Charge: every species in the equation is a neutral molecule or formula unit ( $\text{Cu}(s)$ , $\text{HNO}_3(aq)$ , $\text{Cu(NO}_3)_2(aq)$ , $\text{NO}(g)$ , $\text{H}_2\text{O}(l)$ each written as a whole, uncharged unit), so the total charge on both sides is $0 = 0$ ✓.

All atoms and the overall charge balance, confirming the equation is correctly balanced.

Final answers

(a) Cu in $\text{Cu}(s) = 0$ ; N in $\text{HNO}_3(aq) = +5$ ; N in $\text{NO}(g) = +2$ .
(b) Cu oxidised : N reduced $= \boxed{3:2}$ .
(c) $3\text{Cu}(s) + 8\text{HNO}_3(aq) \rightarrow 3\text{Cu(NO}_3)_2(aq) + 2\text{NO}(g) + 4\text{H}_2\text{O}(l)$ ; $6$ of the $8$ N atoms are unchanged spectator $\text{NO}_3^-$ ions, only $2$ are reduced.
(d) Atoms and charge both confirmed balanced (see working above).