Medium
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
(N/A) The Gibbs free energy of formation $\Delta_r G$ of $Cu_2S$ is less negative than that of $H_2S$ and $CS_2$.
Therefore,$H_2$ and $C$ cannot reduce $Cu_2S$ to $Cu$.
On the other hand,the Gibbs free energy of formation of $Cu_2O$ is greater than that of $CO$. Hence,$C$ can reduce $Cu_2O$ to $Cu$.
$C_{(s)} + Cu_2O_{(s)} \longrightarrow 2Cu_{(s)} + CO_{(g)}$
Thus,the extraction of copper from its pyrite ore is more difficult than from its oxide ore through reduction.
Therefore,$H_2$ and $C$ cannot reduce $Cu_2S$ to $Cu$.
On the other hand,the Gibbs free energy of formation of $Cu_2O$ is greater than that of $CO$. Hence,$C$ can reduce $Cu_2O$ to $Cu$.
$C_{(s)} + Cu_2O_{(s)} \longrightarrow 2Cu_{(s)} + CO_{(g)}$
Thus,the extraction of copper from its pyrite ore is more difficult than from its oxide ore through reduction.
Explore More
Similar Questions
In the electrolytic extraction of $Al$,molten cryolite $(Na_3AlF_6)$ is used because it .......
Medium
View SolutionThe value of ${\Delta _r}{G^\Theta }$ for the formation of $Cr_2O_3$ is $-540 \, kJ \, mol^{-1}$ and that of $Al_2O_3$ is $-827 \, kJ \, mol^{-1}$. Is the reduction of $Cr_2O_3$ possible with $Al$?
Medium
View SolutionThe following reactions take place in the blast furnace during the preparation of impure iron. Identify the reaction pertaining to the formation of the slag.
MediumAIPMT 2011
View SolutionIn which of the following pairs of metals are both commercially extracted from their respective ores by the carbon reduction method?
Difficult
View SolutionGiven below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.
Reason $R$: The value of $\Delta G^{\ominus}$ becomes more on negative side as entropy is higher in liquid state than solid state.
In the light of the above statements,choose the most appropriate answer from the options given below.
Assertion $A$: The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.
Reason $R$: The value of $\Delta G^{\ominus}$ becomes more on negative side as entropy is higher in liquid state than solid state.
In the light of the above statements,choose the most appropriate answer from the options given below.
DifficultJEE MAIN 2022
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo