Read the passage given below to answer questions:
A potential difference developed between electrode and electrolyte is called electrode potential. When the concentrations of all the species involved in a half cell is unity, then electrode potential is known as standard electrode potential. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to solution. The other half-cell in which reduction takes place is called cathode and it has a positive potential with respect to solution. Thus, there exists a potential difference between two electrodes, cathode and anode. This difference is called cell potential and is measured in volts. It is called the cell electromotive force when no current is drawn through the cell. A galvanic cell is represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by salt bridge. Under this convention, emf of cell is positive and is given as
Ecell = Eright - Eleft
The standard electrode potential is very important. The value at standard electrode potential of an electrode is greater than zero, then its reduced form is more stable compared to hydrogen gas. The value at some standard electrode potentials at 298 K are given below. (Ions are present as aqueous species and H2O as liquid)
| Eo/V | |
| Ag+/Ag(s) | 0.80 |
| Cu2+/Cu(s) | 0.34 |
| Pb2+/Pb(s) | -0.13 |
| Fe2+/Fe(s) | -0.44 |
| Mg2+/Mg(s) | -2.36 |
The strongest oxidising agent amongst the following Ag+, Cu2+, Fe2+, Pb2+ is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Ag+
The higher the standard reduction potential of a couple, the greater the tendency of its oxidised form to gain electrons, i.e. the stronger it is as an oxidising agent.
From the table: \( \ce{Ag^{+}} \)/Ag 0.80 V > \( \ce{Cu^{2+}} \)/Cu 0.34 V > \( \ce{Pb^{2+}} \)/Pb −0.13 V > \( \ce{Fe^{2+}} \)/Fe −0.44 V. Oxidising power: Ag+ > \( \ce{Cu^{2+}} \) > \( \ce{Pb^{2+}} \) > \( \ce{Fe^{2+}} \).
Why the other options are incorrect
Option (2) Cu2+ — Incorrect. E° = 0.34 V, lower than that of Ag+/Ag.
Option (3) Fe2+ — Incorrect. E° = −0.44 V, the lowest of the four, so the weakest oxidising agent.
Option (4) Pb2+ — Incorrect. E° = −0.13 V, much lower than 0.80 V.
NCERT (2026-27 edition):
Measurement of Electrode Potential – “… If the standard electrode potential of an electrode is greater than zero then its reduced form is more stable compared to hydrogen gas. Similarly, if the standard electrode potential is negative then hydrogen gas is more stable than the reduced form of the species. It can be seen that the standard electrode potential for fluorine is the highest in the Table indicating that fluorine gas (F2) has the maximum tendency to get reduced to fluoride ions (F–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution.”