Explanation:

Copper has its standard oxidation potential below \(\mathrm{H}\) and thus cannot liberate \(\mathrm{H}_{2}\) from \(\mathrm{HCl}\). However, it dissolves in concentrated nitric acid because conc. \(\mathrm{HNO}_{3}\) is strong oxidising agent.
\(\mathrm{Cu}+4 \mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NO}_{2}\)
So option (1) is correct.

Explanation:

\({\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + KOH}} \to \mathop {{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}}\limits_{{\text{(dark}}\;{\text{green)}}} \xrightarrow[{{\text{solution}}}]{{{\text{Acidic}}}}\mathop {{\text{KMn}}{{\text{O}}_{\text{4}}}}\limits_{{\text{(dark}}\;{\text{purple)}}} \)

Explanation:

\(\mathrm{\mathrm{Cu}+2 \mathrm{AgNO}_{3} \rightarrow 2 \mathrm{Ag}+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}}\)
\(\mathrm{\mathrm{Cu}+\mathrm{ZnSO}_{4} \rightarrow}\) no reaction
\(\mathrm{\mathrm{Cu}+\mathrm{ZnSO}_{4} \rightarrow}\) no reaction
\(\mathrm{E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{~V} \quad E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=+0.34 \mathrm{~V}}\)
\(\mathrm{E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{0}=-0.40 \mathrm{~V} \quad \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V}}\)
As it is clear that reduction potential of copper is more than \(\mathrm{\mathrm{Zn}}\) and Fe. Hence, it is unable to displace them from their salts.

Explanation:

Explanation:

Copper has its standard oxidation potential below \(\mathrm{H}\) and thus cannot liberate \(\mathrm{H}_{2}\) from \(\mathrm{HCl}\). However, it dissolves in concentrated nitric acid because conc. \(\mathrm{HNO}_{3}\) is strong oxidising agent.
\(\mathrm{Cu}+4 \mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NO}_{2}\)
So option (1) is correct.

Explanation:

\({\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + KOH}} \to \mathop {{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}}\limits_{{\text{(dark}}\;{\text{green)}}} \xrightarrow[{{\text{solution}}}]{{{\text{Acidic}}}}\mathop {{\text{KMn}}{{\text{O}}_{\text{4}}}}\limits_{{\text{(dark}}\;{\text{purple)}}} \)

Explanation:

\(\mathrm{\mathrm{Cu}+2 \mathrm{AgNO}_{3} \rightarrow 2 \mathrm{Ag}+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}}\)
\(\mathrm{\mathrm{Cu}+\mathrm{ZnSO}_{4} \rightarrow}\) no reaction
\(\mathrm{\mathrm{Cu}+\mathrm{ZnSO}_{4} \rightarrow}\) no reaction
\(\mathrm{E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{~V} \quad E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=+0.34 \mathrm{~V}}\)
\(\mathrm{E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{0}=-0.40 \mathrm{~V} \quad \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V}}\)
As it is clear that reduction potential of copper is more than \(\mathrm{\mathrm{Zn}}\) and Fe. Hence, it is unable to displace them from their salts.

Explanation:

Explanation:

Copper has its standard oxidation potential below \(\mathrm{H}\) and thus cannot liberate \(\mathrm{H}_{2}\) from \(\mathrm{HCl}\). However, it dissolves in concentrated nitric acid because conc. \(\mathrm{HNO}_{3}\) is strong oxidising agent.
\(\mathrm{Cu}+4 \mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NO}_{2}\)
So option (1) is correct.

Explanation:

\({\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + KOH}} \to \mathop {{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}}\limits_{{\text{(dark}}\;{\text{green)}}} \xrightarrow[{{\text{solution}}}]{{{\text{Acidic}}}}\mathop {{\text{KMn}}{{\text{O}}_{\text{4}}}}\limits_{{\text{(dark}}\;{\text{purple)}}} \)

Explanation:

\(\mathrm{\mathrm{Cu}+2 \mathrm{AgNO}_{3} \rightarrow 2 \mathrm{Ag}+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}}\)
\(\mathrm{\mathrm{Cu}+\mathrm{ZnSO}_{4} \rightarrow}\) no reaction
\(\mathrm{\mathrm{Cu}+\mathrm{ZnSO}_{4} \rightarrow}\) no reaction
\(\mathrm{E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{~V} \quad E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=+0.34 \mathrm{~V}}\)
\(\mathrm{E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{0}=-0.40 \mathrm{~V} \quad \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V}}\)
As it is clear that reduction potential of copper is more than \(\mathrm{\mathrm{Zn}}\) and Fe. Hence, it is unable to displace them from their salts.

Explanation:

Explanation:

Copper has its standard oxidation potential below \(\mathrm{H}\) and thus cannot liberate \(\mathrm{H}_{2}\) from \(\mathrm{HCl}\). However, it dissolves in concentrated nitric acid because conc. \(\mathrm{HNO}_{3}\) is strong oxidising agent.
\(\mathrm{Cu}+4 \mathrm{HNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NO}_{2}\)
So option (1) is correct.

Explanation:

\({\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + KOH}} \to \mathop {{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}}\limits_{{\text{(dark}}\;{\text{green)}}} \xrightarrow[{{\text{solution}}}]{{{\text{Acidic}}}}\mathop {{\text{KMn}}{{\text{O}}_{\text{4}}}}\limits_{{\text{(dark}}\;{\text{purple)}}} \)

Explanation:

\(\mathrm{\mathrm{Cu}+2 \mathrm{AgNO}_{3} \rightarrow 2 \mathrm{Ag}+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}}\)
\(\mathrm{\mathrm{Cu}+\mathrm{ZnSO}_{4} \rightarrow}\) no reaction
\(\mathrm{\mathrm{Cu}+\mathrm{ZnSO}_{4} \rightarrow}\) no reaction
\(\mathrm{E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{~V} \quad E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=+0.34 \mathrm{~V}}\)
\(\mathrm{E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{0}=-0.40 \mathrm{~V} \quad \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V}}\)
As it is clear that reduction potential of copper is more than \(\mathrm{\mathrm{Zn}}\) and Fe. Hence, it is unable to displace them from their salts.

Explanation: