Question

Most probable oxidation state of $Pb$ and $Sn$ will be:
$Pb{O_2} - - - - - PbO\Delta {G_{298}} < 0$
$Sn{O_2} - - - - - SnO{{ }}\Delta {G_{298}} > 0$
A. $P{b^{ + 4}},S{n^{ + 4}}$
B. $P{b^{ + 4}},S{n^{ + 2}}$
C. $P{b^{ + 2}},S{n^{ + 2}}$
D. $P{b^{ + 2}},S{n^{ + 4}}$

Answer 1

Hint: S is the entropy of the system and is the measure of the randomness while H is called the enthalpy of the reaction. G is called the Gibbs energy and is dependent on both the enthalpy and entropy. Change in free energy is the net energy available to do useful work and is thus a measure of the free energy. For this reason, it is also known as the free energy of the reaction.

Complete step by step answer:
All naturally occurring processes tend to proceed in a single direction spontaneously. Here spontaneity means the potential to proceed without the assistance of external agents. S is called the entropy of the reaction and is the measure of the disorder.
The entropy is usually defined for an isolated system where there is always a tendency for the system to become more disordered. The greater the disorder in an isolated system the higher the entropy.
Neither decrease in enthalpy nor the increase in entropy can determine the spontaneity of the system alone.
$G$ is the Gibbs energy or Gibbs function. It is a path function means its value doesn’t depend on the path followed to reach that result. The Gibbs function has the formula of
$G = H - TS$
Where H is enthalpy and S is entropy whereas T is the temperature.
The change in the Gibbs energy of the system, $\Delta {G_{sys}}$ is written as
$\Delta {G_{sys}} = \Delta {H_{sys}} - T\Delta {S_{sys}} - {S_{sys}}\Delta T$
At constant temperature $\Delta T = 0$
Thus, $\Delta G = \Delta H - T\Delta S$ (The subscript sys is dropped for simplicity)
The above equation is called the Gibbs equation. This equation takes into account both the entropy and the enthalpy of the system.
For spontaneous process the change in entropy should be greater than 0 in other words
$\Delta S > 0$
In the process of reduction, the change in enthalpy of the system is less than 0 since it is generally an endothermic process.
Using the above points we can infer that for a reduction reaction $\Delta H < 0$ and which upon putting in the Gibbs equation makes $\Delta G < 0$
In the first equation, we see that Gibb’s energy for the conversion is less than zero hence the reaction is stable, so the oxidation state of lead is stable in the conversion that is in the state of +2. While in the second equation we see that Gibb’s energy is positive hence not feasible. So the reaction is not viable so the oxidation state of tin will be +4 as in the actual initial state.

So, the correct answer is Option D.

Note: If the reaction has a positive enthalpy change and positive entropy change, it can be spontaneous only when the effect of $T\Delta S$ outweigh $\Delta H$.
This can be possible when the positive entropy change of the system is small and T must be large or the positive entropy change of the system may be large and T may be small.
When a system is in equilibrium, the entropy is maximum and thus the change in the entropy is 0
Thus, the entropy of the spontaneous process increases till it reaches the maximum and the equilibrium is reached.