Carbon Free energy vs Temperature Diagram

Question

I've recently read that the Carbon takes a unique position as a reducing agent , and i think it has to do with the very special form of its Ellingham diagram , but i am still not able to make a connection , So what makes exaclty the carbon a good reducing agent ?

Answer 1

Here is the link to a set of Ellingham Diagrams

At high temperatures, smelting and carbon reductions have several practical challenges some of which are energy efficiency, formation of metal carbides, equipment concerns etc. Therefore, to assess the reducing ability of any reducing agent, we must work within practical limits. Let us select a temperature of $1600\ \mathrm{K}$ as our upper bound beyond (which is still quite high) which, practical concerns outweigh any benefit of having a good reductant. At this temperature $\Delta_\text fG(\ce{CO2})$ is approx $-394.39\ \mathrm{kJ/mol}$ and $\Delta_\text fG(\ce{CO})$ is $-504\ \mathrm{kJ/mol}$. These values are high enough and can already reduce several oxides of metals like $\ce{Co, Fe, Cu, Pb, Ni, Bi}$ and $\ce{MnO2, CrO2}$ etc (subject to practical constraints). This gives it a good reducing character, better than $\ce{H2}$ or ammonia.

Another thing to notice, is that, although the variation of $\Delta_\text fG$ for carbon dioxide is flat, for carbon monoxide it slopes downward and hence with increasing temperature, it can reduce many more oxides further expanding its capacity to reduce metal oxides. $\ce{V2O5, VO2, SnO2, WO3, Mn3O4, MoO3, GeO2}$ can be reduced by carbon. (All this just implies thermodynamic favourability; it can still be unfavourable kinetically, or due to practical restraints.) Hence, the negative sloping $\Delta_\text fG$ for $\ce{CO}$ makes carbon a very good reducing agent, and it can reduce a huge variety of metal oxides, within a considerable temperature limit.