The choice of a reducing agent in a particular case depends on thermodynamic factors. How far do you agree with this statement? Support your opinion with two examples.

(N/A) Yes,the choice of a reducing agent is primarily governed by thermodynamic factors,specifically the Gibbs free energy change $(\Delta G^{\Theta})$. $A$ metal can reduce the oxide of another metal if the standard free energy of formation $(\Delta_f G^{\Theta})$ of the oxide of the reducing metal is more negative than that of the metal being reduced.
Example $1$: Aluminum $(Al)$ can reduce Copper$(I)$ oxide $(Cu_2O)$ to Copper $(Cu)$ because $\Delta_f G^{\Theta} (Al_2O_3)$ is more negative than $\Delta_f G^{\Theta} (Cu_2O)$. The reaction is: $2Al + 3Cu_2O \rightarrow Al_2O_3 + 6Cu$.
Example $2$: Magnesium $(Mg)$ can reduce Zinc oxide $(ZnO)$ to Zinc $(Zn)$ because $\Delta_f G^{\Theta} (MgO)$ is more negative than $\Delta_f G^{\Theta} (ZnO)$. The reaction is: $Mg + ZnO \rightarrow MgO + Zn$.