Question

Cu(s) +NO${}_3^{-}$(aq) +H${}^{+}$(aq) $\to$Cu${}^{2+}$(aq) +NO${}_2$(g) +H${}_2$O(l)
If the above equation is balanced with lowest whole number coefficients, find the coefficient for H+(aq).

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Answer 1

Answer: (D)

4

The unbalanced chemical equation is $Cu\left(s\right)+N{O}_3^{-}\left(aq\right)+H^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+N{O}_2\left(g\right)+H_{2}O\left(l\right)$.
Cu and N atoms are balanced.
The oxidation number of Cu increases from 0 to +2. Total increase in the oxidation number is 2.
The oxidation number of N decreases from +5 to +4. Total decrease in oxidation number is 1.
Balance increase in oxidation number with decrease in oxidation number by multiplying the N species with 2.
$Cu\left(s\right)+2N{O}_3^{-}\left(aq\right)+H^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2N{O}_2\left(g\right)+H_{2}O\left(l\right)$
Balance O atoms by adding one water molecule on RHS.
$Cu\left(s\right)+2N{O}_3^{-}\left(aq\right)+H^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2N{O}_2\left(g\right)+2H_{2}O\left(l\right)$
Balance H atoms by adding coefficient 4 to proton on LHS.
$Cu\left(s\right)+2N{O}_3^{-}\left(aq\right)+4H^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2N{O}_2\left(g\right)+2H_{2}O\left(l\right)$
This is the balanced chemical equation. The coefficient for $H^{+}$ is 4.