$\Delta {{\text{G}}^ \circ }$ versus T plot in the Ellingham’s diagram slopes downward for which of the reaction?
A. ${\text{C}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{CO}}$
B.${\text{CO}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{C}}{{\text{O}}_2}$
C.${\text{2}}\,{\text{Ag + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{A}}{{\text{g}}_2}{\text{O}}$
D. ${\text{Mg}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{MgO}}$

Hint: Ellingham diagram is used to determine that a reduction process of ore will be feasible or not and by the help of the Ellingham diagram, a reducing agent is also selected for the reduction of ore.

Complete step by step answer:
A graphical representation of the change in Gibbs free energy and temperature is shown by the Ellingham diagram.
Ellingham diagram is used to select a reducing agent and the feasibility of the reduction process.
For a reaction to be feasible the change in Gibbs free energy \[\Delta {\text{G}}\] should be negative.
The relation between Gibbs free erg change and the temperature is as follows:
$\Delta {\text{G = }}\Delta {\text{H}}\, - {\text{T}}\Delta {\text{S}}$
If ${\text{T}}\Delta {\text{S}}\,{\text{ > }}\Delta {\text{H}}$, the change in Gibbs free energy will be negative.
If ${\text{T}}\Delta {\text{S}}\,{\text{ < }}\Delta {\text{H}}$, the change in Gibbs free energy will be positive.
Features of Ellingham diagram: During the reaction the gas is utilized so, the entropy becomes negative which leads to the positive value of the term,${\text{T}}\Delta {\text{S}}\,$. So, slope $\Delta {\text{S}}$ comes positive for most of the reduction reaction.
For ${\text{C}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{CO}}$, the oxygen gas is consuming but the formed product is also a gas so, the number of gaseous species in increasing hence entropy increasing so, for the oxidation reaction of carbon to carbon monoxide the slope is downward. So, option (A) is correct.
For ${\text{CO}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{C}}{{\text{O}}_2}$, the oxygen gas is consuming and the formed product is also a gas but the total number of gaseous species is decreasing so, the entropy decreasing so, for the oxidation reaction of carbon monoxide to carbon dioxide the slope is upward. So, option (B) is incorrect.
For ${\text{2}}\,{\text{Ag + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{A}}{{\text{g}}_2}{\text{O}}$, the oxygen gas is consuming and the formed product is solid so, for the oxidation reaction of silver to silver oxide the slope is upward. So, option (C) is incorrect.
For ${\text{Mg}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{MgO}}$, the oxygen gas is consuming and the formed product is solid so, for the oxidation reaction of magnesium to magnesium oxide the slope is upward. So, option (D) is incorrect.

Therefore, option (A) ${\text{C}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\, \to \,{\text{CO}}$ is correct.

Note: The Ellingham diagram is based on the thermodynamic criteria only for the feasibility of a reduction reaction. Some other factors also affect the reaction such as kinetics. Ellingham diagram is based upon the assumption that the reactant and product are in equilibrium.