Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below

Q: Why do we multiply the dissociation enthalpy by 1/2?

The given dissociation enthalpy is for one full mole of $Cl_2$ molecules (producing 2 moles of $Cl$ atoms). Since our target reaction only requires one mole of $Cl$ atoms, we only need half a mole of $Cl_2$. Therefore, we take half the energy.

Q: Why are the electron gain and hydration enthalpies negative?

A negative sign indicates an exothermic process, meaning energy is released. Chlorine has a high affinity for electrons, so gaining one releases energy. Similarly, water molecules strongly attract the charged $Cl^-$ ion, releasing hydration energy.

Q: What does the final negative $\Delta H$ tell us about chlorine?

The large negative value ($-610 \text{ kJ mol}^{-1}$) means the overall process is highly exothermic and thermodynamically favorable. This massive release of energy is the fundamental reason why chlorine is such a strong oxidizing agent in aqueous solutions.

Q: Can we use this same cycle for other halogens like fluorine or bromine?

Absolutely! This is a standard Born-Haber-type cycle used for all halogens. In fact, comparing the individual terms (like dissociation vs. hydration) explains why fluorine is an even stronger oxidizing agent than chlorine despite having a lower electron gain enthalpy.

Q: Is the oxidizing power only dependent on enthalpy?

Technically, spontaneity and oxidizing power depend on the Gibbs Free Energy change ($\Delta G$). However, for these reactions, the enthalpy change ($\Delta H$) is so large that it dominates the entropy term ($T\Delta S$), making $\Delta H$ a very reliable proxy for oxidizing power.

Question

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below
$\frac{1}{2}Cl_2(g) \xrightarrow{\frac{1}{2}\Delta_{diss}H^{\ominus}} Cl(g) \xrightarrow{\Delta_{EA}H^{\ominus}} Cl^-(g) \xrightarrow{\Delta_{hyd}H^{\ominus}} Cl^-(aq)$
The energy involved in the conversion of $\frac{1}{2}Cl_2(g)$ to $Cl^-(aq)$ (using the data, $\Delta_{diss}H^{\ominus}_{Cl_2} = 240 	ext{ kJ mol}^{-1}$, $\Delta_{EA}H^{\ominus}_{Cl} = -349 	ext{ kJ mol}^{-1}$, $\Delta_{hyd}H^{\ominus}_{Cl^-} = -381 	ext{ kJ mol}^{-1}$) will be

Options:

(A) $+152 	ext{ kJ mol}^{-1}$

(B) $-610 	ext{ kJ mol}^{-1}$

(C) $-850 	ext{ kJ mol}^{-1}$

(D) $+120 	ext{ kJ mol}^{-1}$

Sigma Percentile · Accepted Answer

### The Anatomy of Oxidizing Power

Have you ever wondered what makes a chemical a strong oxidizing agent? In the realm of chemistry, an oxidizing agent is a substance that forces another substance to lose electrons while it happily gains them. But this electron theft isn't just a simple transaction; it's a thermodynamic journey.

To truly understand the oxidizing power of a halogen like chlorine in an aqueous solution, we must break down its transformation from a diatomic gas into a hydrated ion. This is where the elegance of **Hess's Law** and the **Born-Haber cycle** comes into play. Hess's Law tells us that the total enthalpy change of a reaction is simply the sum of the enthalpy changes of its individual steps, regardless of the path taken.

### Step 1: Breaking the Bond (Atomization)

Our journey begins with a stable, diatomic chlorine molecule, $Cl_2(g)$. Before it can steal any electrons, it must be broken apart into individual atoms. This process requires an input of energy, known as the bond dissociation enthalpy ($\Delta_{diss}H^{\ominus}$).

However, there is a crucial catch here. The standard dissociation enthalpy is given for breaking one full mole of $Cl_2$ molecules, which would yield two moles of $Cl$ atoms. But our target reaction only requires *one* mole of $Cl$ atoms. Therefore, we must take exactly half of the dissociation energy:

$\Delta H_1 = \frac{1}{2} \Delta_{diss}H^{\ominus} = \frac{240}{2} = +120 	ext{ kJ mol}^{-1}$

This step is **endothermic**; we are investing energy into the system to break the bond.

### Step 2: The Electron Grab (Electron Gain Enthalpy)

Now we have a highly reactive, gaseous chlorine atom, $Cl(g)$. Chlorine is a halogen, meaning it is just one electron shy of a stable noble gas configuration. When it encounters an electron, it grabs it aggressively, forming a chloride ion, $Cl^-(g)$.

Because chlorine strongly desires this electron, the process releases a significant amount of energy. This is the electron gain enthalpy ($\Delta_{EA}H^{\ominus}$):

$\Delta H_2 = \Delta_{EA}H^{\ominus} = -349 	ext{ kJ mol}^{-1}$

The negative sign indicates that this step is highly **exothermic**.

### Step 3: The Water Embrace (Hydration Enthalpy)

Our chloride ion is now negatively charged, but it is still in the gaseous state. In an aqueous solution, it gets dropped into water. Water molecules are polar; their slightly positive hydrogen ends are strongly attracted to the negative chloride ion.

As water molecules surround and stabilize the ion, a massive amount of energy is released. This is the hydration enthalpy ($\Delta_{hyd}H^{\ominus}$):

$\Delta H_3 = \Delta_{hyd}H^{\ominus} = -381 	ext{ kJ mol}^{-1}$

This step is also highly **exothermic** and plays a massive role in driving the overall reaction forward.

### The Master Equation: Summing It Up

Now, we invoke Hess's Law to find the total energy change for the entire process. We simply add the enthalpies of the three steps together:

$\Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3$

Substituting our values:

$\Delta H_{total} = 120 + (-349) + (-381)$

$\Delta H_{total} = 120 - 730$

$\Delta H_{total} = -610 	ext{ kJ mol}^{-1}$

### The Physical Significance

The final result is a staggering **$-610 	ext{ kJ mol}^{-1}$**. This large negative value means that the overall transformation of gaseous chlorine into aqueous chloride ions releases a tremendous amount of energy.

In thermodynamics, systems naturally progress toward lower energy states. Because this process is so highly exothermic, chlorine has a very strong thermodynamic drive to undergo this transformation. This immense drive to gain an electron and become hydrated is the exact physical reason why chlorine is such a potent and effective oxidizing agent in aqueous solutions.

Select Answer:

Visualized Solution

\text{The Overall Process}

\text{Hess's Law}

\text{Step 1: Dissociation of } Cl_2

\text{Step 2: Electron Gain Enthalpy}

\text{Step 3: Hydration Enthalpy}

\text{Substituting Values}

\text{Calculation}

\text{Final Result}

\text{Conclusion}

The Sigma Insight: Enthalpy and Hess's Law

The Anatomy of Oxidizing Power

Step 1

Step 2

Step 3

The Master Equation

The Physical Significance

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