Arrange the following redox couple in increasing order of their oxidising strength. The standard electrode potential of the redox couples are given in table.
| Redox couple | Standard electrode potential (Xn+/X) |
| (A) | 0.79 V |
| (B) | 0.20 V |
| (C) | -0.14 V |
| (D) | -0.34 V |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (D), (C), (B), (A)
The standard electrode (reduction) potential E°(Xn+/X) measures the tendency of the oxidised form to gain electrons:
The more positive the E°, the more easily Xn+ is reduced, so the stronger it is as an oxidising agent.
The more negative the E°, the weaker the oxidising strength (and the stronger X is as a reducing agent).
Arranging the given couples from the most negative to the most positive E°:
−0.34 V < −0.14 V < 0.20 V < 0.79 V
(D) < (C) < (B) < (A)
Shortcut: The couple with the most negative E° (D) is the weakest oxidant and the most positive one (A) the strongest. Only Option (4) starts with (D) and ends with (A).
NCERT (2026-27 edition):
Measurement of Electrode Potential – “… It may be seen that as we go from top to bottom in Table 2.1 the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction.”