Question

When the following redox reaction
$Cu\left(aq\right)+N{O}_3^{-}\left(aq\right)\to C{u}^{2+}\left(aq\right)+N{O}_2\left(g\right)$ is balanced by oxidation number method in acidic meduim, the stoichiometric ratio $\frac{n_{N{O}_2}}{n_{H_{2}O}}$ will be :

• A. 0.5
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

$Cu\left(aq\right)+N{O}_3^{-}\left(aq\right)\to C{u}^{2+}\left(aq\right)+N{O}_2\left(g\right)$

Here oxidation number of $Cu$ increases from $0$ to $+2$
$Cu$ is the reducing agent.
$n_f=2$

Oxidation number of $N$ decreases from $+5$ to $+4$
$N{O}_3^{-}$ is the oxidising agent.
$n_f=1$

Cross multiplying $n_f$ values
$Cu\left(aq\right)+2N{O}_3^{-}\left(aq\right)\to C{u}^{2+}\left(aq\right)+N{O}_2\left(g\right)$

Balancing all elements except $O$ and $H$
$Cu\left(aq\right)+2N{O}_3^{-}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2N{O}_2\left(g\right)$

Balancing oxygen by adding $H_{2}O$

$Cu\left(aq\right)+2N{O}_3^{-}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2N{O}_2\left(g\right)+2H_{2}O$

Balancing hydrogen by adding $H^{+}$

$Cu\left(aq\right)+2N{O}_3^{-}\left(aq\right)+4H^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2N{O}_2\left(g\right)+2H_{2}O\left(l\right)$
Charges on both reactant and product side are $+2$ and are equal to each other.
So, the above reaction is balanced.
The stoichiometric ratio :
$\frac{n_{N{O}_2}}{n_{H_{2}O}}=\frac{2}{2}=1$