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

The above figure is a plot of Gibbs energy  $\left(∆{G}^{\theta }\right)$ vs. T for formation of some oxides. It can be observed from the above graph that metal can reduce the oxide of other metals, if the standard free energy of formation  $\left({∆}_{t}{G}^{\theta }\right)$ of the oxide of the former is more negative than the latter. For example, since ${∆}_{t}{{G}^{\theta }}_{\left(AI,A{I}_{2},{O}_{3}\right)}$ is more negative than , ${∆}_{t}{{G}^{\theta }}_{\left(Cu,C{u}_{2},O\right)}$ Al can reduce Cu2O to Cu, but Cu cannot reduce $A{I}_{2}{O}_{3}$. Similarly, Mg can reduce ZnO to Zn, but Zn cannot reduce MgO because ${∆}_{t}{{G}^{\theta }}_{\left(Mg,MgO\right)}$ is more negative than ${∆}_{t}{{G}^{\theta }}_{\left(Zn,ZnO\right)}$

In the electrolysis of molten NaCl, $C{L}_{2}$ is obtained at the anode as a by product.
If an aqueous solution of NaCl is electrolyzed, $C{l}_{2}$ will be obtained at the anode but at the cathode, ${H}_{2}$ will be obtained (instead of Na). This is because the standard reduction potential of Na (E°= − 2.71 V) is more negative than that of ${H}_{2}O$ (E° = − 0.83 V). Hence, ${H}_{2}O$ will get preference to get reduced at the cathode and as a result, ${H}_{2}$ is evolved.