Match List-I with List-II
| List-I | List-II |
| Parameters | Characteristics |
| (A) Conductivity | (I) increases with dilution |
| (B) Molar conductivity | (II) Redox couple is a stronger oxidizing agent than H+/H2 couple |
| (C) Negative standard reduction potential | (III) decreases with dilution |
| (D) Positive standard reduction potential | (IV) Redox couple is a stronger reducing agent than H+/H2 couple |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Conductivity refers to 1 cm3 of solution, while molar conductivity refers to all the solution that contains 1 mole of electrolyte, so dilution affects them in opposite ways. A standard reduction potential is measured against the \( \ce{H^{+}} \)/\( \ce{H2} \) couple, taken as zero, so its sign shows whether a couple is a stronger or weaker oxidising agent than hydrogen ions.
(A) Conductivity — This matches (III) decreases with dilution. Conductivity is the conductance of 1 cm3 of solution. On dilution, fewer ions are present in each cm3, so conductivity falls.
(B) Molar conductivity — This matches (I) increases with dilution. Molar conductivity is the conductance of the whole volume that contains 1 mole of electrolyte. On dilution, this volume increases much more than the conductivity falls, and the ions move more freely (strong electrolytes) or dissociate more (weak electrolytes).
\( \displaystyle \Lambda_m=\frac{\kappa}{c} \)
(C) Negative standard reduction potential — This matches (IV) Redox couple is a stronger reducing agent than H+/H2 couple. A negative E° means the reduced form gives up electrons more easily than \( \ce{H2} \), for example \( \ce{Zn^{2+}} \)/Zn (−0.76 V).
(D) Positive standard reduction potential — This matches (II) Redox couple is a stronger oxidizing agent than H+/H2 couple. A positive E° means the oxidised form accepts electrons more easily than \( \ce{H^{+}} \), for example \( \ce{Cu^{2+}} \)/Cu (+0.34 V).