Question

For $Cu\left(s\right)+N{O}_3^{-}\left(aq\right)+H^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+N{O}_2\left(g\right)+H_{2}O\left(l\right)$, which of the following takes place?

• A. $Cu\left(s\right)$ is oxidized
• B. $H^{+}\left(aq\right)$ is oxidized
• C. $Cu\left(s\right)$ is reduced
• D. $H^{+}\left(aq\right)$ is reduced
• E. $N{O}_3^{-}$ is oxidized.

Answer 1

Answer: (A)

$Cu\left(s\right)$ is oxidized

Oxidation state of $Cu$ changes from $0$ to $+2$

Oxidation state of $N$ changes from $+5$ to $+4$

Oxidation state of $H^{+}$ remains same

therefore, $Cu$ is getting oxidised and $N{O}_3^{-}$ is getting reduced.