
Out of C and CO, which is a better reducing agent for ZnO?
Answer
529.8k+ views
Hint: Reducing agent is a compound or an element that can reduce the other reacting species by donating hydrogen atoms or removing oxygen atoms or by adding electrons to it.
Complete step by step answer:
Carbon and carbon monoxide, both are having good reducing properties. So they act as reducing agents in many reduction reactions. In this particular case of reducing zinc oxide (ZnO), with C or CO is well explained in the thermodynamic point of view given by the Ellingham diagram.
From the plot, for zinc oxide, Gibb’s free energy of the formation of carbon dioxide from carbon is lesser than Gibbs’s energy of the formation of zinc oxide. And Gibb’s energy of the formation of carbon dioxide from carbon monoxide is higher than that of the formation of zinc oxide. From these observations, it can be understood that carbon in its elemental form (C) can reduce the metal oxide more effectively than carbon monoxide. This is because, less the Gibbs energy, more is the feasibility of the reaction as we know$\Delta {{G}^{0}}<0$, or negative the reaction is spontaneous.
Therefore the answer for the above question out of C and CO, which is a better reducing agent for ZnO, is ‘C’.
${\text{ZnO + C}}\xrightarrow{{{\text{1676K}}}}{\text{Zn + CO}}$
Note: Ellingham diagram is a graphical plot of Gibbs free energy $\left( {{G}^{0}} \right)$of oxygen vs. temperature of oxide formation which is used in metallurgical application of different reducing agents. From the values of Gibbs energy and temperature, we can know the best-reducing agents for the given oxide or sulfide ores of different metals.
Complete step by step answer:
Carbon and carbon monoxide, both are having good reducing properties. So they act as reducing agents in many reduction reactions. In this particular case of reducing zinc oxide (ZnO), with C or CO is well explained in the thermodynamic point of view given by the Ellingham diagram.
From the plot, for zinc oxide, Gibb’s free energy of the formation of carbon dioxide from carbon is lesser than Gibbs’s energy of the formation of zinc oxide. And Gibb’s energy of the formation of carbon dioxide from carbon monoxide is higher than that of the formation of zinc oxide. From these observations, it can be understood that carbon in its elemental form (C) can reduce the metal oxide more effectively than carbon monoxide. This is because, less the Gibbs energy, more is the feasibility of the reaction as we know$\Delta {{G}^{0}}<0$, or negative the reaction is spontaneous.
Therefore the answer for the above question out of C and CO, which is a better reducing agent for ZnO, is ‘C’.
${\text{ZnO + C}}\xrightarrow{{{\text{1676K}}}}{\text{Zn + CO}}$
Note: Ellingham diagram is a graphical plot of Gibbs free energy $\left( {{G}^{0}} \right)$of oxygen vs. temperature of oxide formation which is used in metallurgical application of different reducing agents. From the values of Gibbs energy and temperature, we can know the best-reducing agents for the given oxide or sulfide ores of different metals.
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