question_answer 25)

 Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration energy, compare the oxidizing power of F2 and Cl2.

Answer:

Electrode potential depends upon three factors: (i) bond dissociation energy, (ii) electron gain enthalpy and (iii) hydration enthalpy. Although electron gain enthalpy of fluorine is less negative (–333 kJ mol–1) than that of chlorine (–349 kJ mol–1), the bond dissociation energy of F – F bond is much lower (158.5 kJ mol–1) than that of Cl – Cl bond (242.6 kJ mol–1) and hydration enthalpy of F– ions (515 kJ mol–1) is much higher than that of Cl– ion (381 kJ mol–1). The later two factors more than compensate the less negative electron gain enthalpy of F2. As a result, electrode reduction potential of F2(+2.87V) is higher than that of chlorine (+1.36V). Thus F2 is a much stronger oxidizing agents than Cl2.