Based on the standard electrode potential given in the table

Choose the wrong/incorrect statement from the options given below:

Lithium ion is a strong oxidizing agent.

The correct answer is Option (2) → Lithium ion is a strong oxidizing agent.

Analysis of the Statements

Fluorine gas is a strong oxidizing agent: CORRECT.

$F_2$ has the highest positive reduction potential in the table ($+2.87\text{ V}$). This means it has the strongest tendency to gain electrons and undergo reduction, making it a very powerful oxidizing agent.

Lithium ion is a strong oxidizing agent: INCORRECT.

$Li^+$ has the most negative reduction potential ($-3.05\text{ V}$). This means the $Li^+$ ion is extremely difficult to reduce. In contrast, Lithium metal ($Li$) is a very strong reducing agent, but the Lithium ion ($Li^+$) is a very weak oxidizing agent.

Copper can be oxidized by nitrate ion: CORRECT.

For a reaction to be spontaneous, the oxidizing agent must have a higher $E^\circ$ than the substance being oxidized. The nitrate ion ($NO_3^-$) has an $E^\circ$ of $+0.97\text{ V}$, which is higher than the $E^\circ$ for $Cu^{2+}/Cu$ ($+0.34\text{ V}$). Therefore, $NO_3^-$ can spontaneously oxidize $Cu$ to $Cu^{2+}$.

Zinc can be oxidized by hydrogen ion: CORRECT.

The $H^+/H_2$ couple has an $E^\circ$ of $0.00\text{ V}$. Zinc ($Zn^{2+}/Zn$) has an $E^\circ$ of $-0.76\text{ V}$. Since $0.00 > -0.76$, the hydrogen ion ($H^+$) can spontaneously oxidize Zinc metal ($Zn$) to $Zn^{2+}$.