Statement I: Cu is less reactive than hydrogen.

Statement II: \(E_{Cu^{2+}/Cu}\) is negative.

Statement I is correct but Statement II is false

The correct answer is option 3. Statement I is correct but Statement II is false.

Let us delve into both the statements and their implications:

Statement I: Cu is less reactive than hydrogen

Statement I is correct because copper is indeed less reactive than hydrogen.

In the reactivity series of metals, copper \((Cu)\) is placed below hydrogen. This means copper is less reactive than hydrogen. A more reactive metal can displace a less reactive metal from its compounds. Since copper cannot displace hydrogen from acids (such as \(HCl\) or \(H_2SO_4\)), it shows that copper is less reactive than hydrogen. Metals with more positive standard electrode potentials are less reactive because they have a greater tendency to gain electrons. Copper has a more positive electrode potential compared to hydrogen, reinforcing its lower reactivity.

Statement II: \(E^\circ_{Cu^{2+}/Cu}\) is negative

Statement II is false because the standard electrode potential for the copper(II)/copper couple is positive, not negative.

The standard reduction potential for the reaction \(Cu^{2+} + 2e^- \rightarrow Cu\) is +0.34 V. This positive value indicates that \(Cu^{2+}\) ions are more likely to gain electrons and be reduced to metallic copper.

A negative \(E^\circ\) would imply that the reduction of \(Cu^{2+}\) to Cu is not favorable under standard conditions, which is not the case. The positive electrode potential shows that the reduction is thermodynamically favorable.