Question

Name Question/6 Why is F2 a stronger oxidant than Cl2 under standard conditions? Answer by constructing a thermodynamic (or Hess) cyele (write out the equations generically using X to represent the halogen) and justifying the relative values for F vs. Cl in each. Some important information includes ?IENF)=-328 kJ/rmol. ?{ENC,--349 kJ/mol, E.F2F--2.87, E"coor-1.36. Half-cell reaction: Thermodynamic cycle component 1 Thermodynamic cycle component 2: Thermodynamic cycle component 2: Final justification: 41 CHAT

Answer 1

Half Cell Reaction:

X₂ +2e^-$\rightarrow$2X^-

Thermodynamic cycle component 1:

X2 (g) $\rightarrow$ 2X(g) ........$\Delta$H_atomization

X2 has to be supplied bond dissociation energy to form atoms. This is called $\Delta$ H_atomization which is a positive quantity.

Thermodynamic cycle component 2:

X(g) + e^-$\rightarrow$ X^-(g)   ........$\Delta$H(EA)

The gaseous X atoms takes up an electron by releasing energy. The enthalpy change in this process is called electron affinity (EA) which is negative as energy is released in the process.

Thermodynamic cycle component 3:

X^-(g)  $\rightarrow$  X₂(g) ........$\Delta$H(X^-/X₂)

This is the direct conversion of X^-(g) to X₂(g)

From thermodynamics we know, $\Delta$ H(X^-/X₂) = - $\Delta$ H(X₂/X^-)

From the three thermodynamic cycle components and applying Hess law we will get,

$\Delta$H_atomization + $\Delta$ H(EA) + $\Delta$ H(X^-/X₂) = 0

$\Delta$H_atomization + $\Delta$ H(EA) = - $\Delta$ H(X^-/X₂) =  $\Delta$H(X₂/X^-)

$\Delta$H_atomization value of Cl is around 120 kJ/mol while for F it is 79 kJ/mole

$\Delta$H(F₂/F^-) = -328 + 79 = -249 kJ/mol

$\Delta$H(Cl₂/Cl^-) = -349 + 120= -229 kJ/mol

We know $\Delta$ G = $\Delta$ H- T$\Delta$S

If we think $\Delta$ S is similar for both Cl2 and F2 then $\Delta$ G = $\Delta$ H

For Fluorine the value of $\Delta$ G will be more negative and the oxidation process by it will be more spontaneous.