Out of the following half cells, which two half cell combination gives the largest \(E_{cell}\) value?
| \(A^{3-} \longrightarrow A^{2-} + e^-\) | \(E^o = + 1.5 V\) -------(I) |
| \(B^+ + e^- \longrightarrow B\) | \(E^o = - 0.5 V\) -------(II) |
| \(C^{2+} + e^- \longrightarrow C^+\) | \(E^o = + 0.5 V\) -------(III) |
| \(D \longrightarrow D^{2+} + 2e^-\) | \(E^o = - 1.5 V\) -------(IV) |
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → I and IV
Standard electrode potentials are reduction potentials of the redox couples. In a cell, the couple with the higher E° acts as the cathode (reduction) and the one with the lower E° as the anode (oxidation):
E°cell = E°cathode − E°anode
So E°cell is largest when the couple with the highest E° is combined with the couple with the lowest E°. Here the highest value is +1.5 V (I) and the lowest is −1.5 V (IV):
E°cell = (+1.5) − (−1.5) = 3.0 V
Why the other options are incorrect
Option (1) I and III — Incorrect. E°cell = 1.5 − 0.5 = 1.0 V.
Option (3) II and IV — Incorrect. E°cell = (−0.5) − (−1.5) = 1.0 V.
Option (4) III and IV — Incorrect. E°cell = 0.5 − (−1.5) = 2.0 V, less than 3.0 V.
NCERT (2026-27 edition):
Galvanic Cells – “… A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge. Under this convention the emf of the cell is positive and is given by the potential of the half-cell on the right hand side minus the potential of the half-cell on the left hand side i.e.,
Ecell = Eright – Eleft”