Based on the standard electrode potential given in the table Half-cel…

Based on the standard electrode potential given in the table

Half-cell reaction	(E°/V) at 25 °C
\( F_2(g) +2e^-→2F^- \)	2.87
\( {NO_3}^- + 4H^++3e^- →NO(g) + 2H_2O \)	0.97
\( Cu^{2+}+2e^-→Cu(s) \)	0.34
\( 2H^++ 2e^-→ H_2(g) \)	0.00
\( Zn^{2+} + 2e^-→Zn(s) \)	-0.76
\( Li^+ + e^-→ Li(s) \)	-3.05

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

Answer & explanation

Correct answer: option 2

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

Standard electrode potentials are written as reductions. The more positive the E°, the more easily the species on the left is reduced, so the stronger it is as an oxidising agent. The more negative the E°, the more easily the species on the right is oxidised, so the stronger it is as a reducing agent.

For \( \ce{Li+ + e- -> Li(s)} \), E° = −3.05 V, the lowest value in the table. Lithium metal is therefore the strongest reducing agent, and the lithium ion has almost no tendency to take an electron back: \( \ce{Li+} \) is the weakest oxidising agent in the table.

Option (1) Fluorine gas is a strong oxidising agent. — A correct statement. \( \ce{F2} \) has the highest E° (2.87 V), so it is the strongest oxidising agent in the table.
Option (2) Lithium ion is a strong oxidizing agent. — The incorrect statement. With E° = −3.05 V, \( \ce{Li+} \) is the weakest oxidising agent in the table; it is lithium metal that is a very strong reducing agent.
Option (3) Copper can be oxidized by nitrate ion. — A correct statement. \( \ce{NO3^-} \) (0.97 V) lies above \( \ce{Cu^2+}/\ce{Cu} \) (0.34 V), so \( E^{\circ}_{\text{cell}} = 0.97 - 0.34 = 0.63 \) V is positive and the reaction is spontaneous:
\( \displaystyle \ce{3Cu + 2NO3^- + 8H+ -> 3Cu^2+ + 2NO + 4H2O} \)
Option (4) Zinc can be oxidized by hydrogen ion. — A correct statement. \( \ce{H+}/\ce{H2} \) (0.00 V) lies above \( \ce{Zn^2+}/\ce{Zn} \) (−0.76 V), so \( E^{\circ}_{\text{cell}} = 0.00 - (-0.76) = 0.76 \) V is positive:
\( \displaystyle \ce{Zn + 2H+ -> Zn^2+ + H2 ^} \)

Based on the standard electrode potential given in the table

Half-cell reaction	(E°/V) at 25 °C
\( F_2(g) +2e^-→2F^- \)	2.87
\( {NO_3}^- + 4H^++3e^- →NO(g) + 2H_2O \)	0.97
\( Cu^{2+}+2e^-→Cu(s) \)	0.34
\( 2H^++ 2e^-→ H_2(g) \)	0.00
\( Zn^{2+} + 2e^-→Zn(s) \)	-0.76
\( Li^+ + e^-→ Li(s) \)	-3.05

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

Answer & explanation

Correct answer: option 2

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

Option (1) Fluorine gas is a strong oxidising agent. — A correct statement. \( \ce{F2} \) has the highest E° (2.87 V), so it is the strongest oxidising agent in the table.
Option (2) Lithium ion is a strong oxidizing agent. — The incorrect statement. With E° = −3.05 V, \( \ce{Li+} \) is the weakest oxidising agent in the table; it is lithium metal that is a very strong reducing agent.
Option (3) Copper can be oxidized by nitrate ion. — A correct statement. \( \ce{NO3^-} \) (0.97 V) lies above \( \ce{Cu^2+}/\ce{Cu} \) (0.34 V), so \( E^{\circ}_{\text{cell}} = 0.97 - 0.34 = 0.63 \) V is positive and the reaction is spontaneous:
\( \displaystyle \ce{3Cu + 2NO3^- + 8H+ -> 3Cu^2+ + 2NO + 4H2O} \)
Option (4) Zinc can be oxidized by hydrogen ion. — A correct statement. \( \ce{H+}/\ce{H2} \) (0.00 V) lies above \( \ce{Zn^2+}/\ce{Zn} \) (−0.76 V), so \( E^{\circ}_{\text{cell}} = 0.00 - (-0.76) = 0.76 \) V is positive:
\( \displaystyle \ce{Zn + 2H+ -> Zn^2+ + H2 ^} \)