Question

The final step in the smelting of

${\mathbf{\text{Cu}}}_{\mathbf{\text{2}}}{\mathbf{\text{S(s) + 2Cu}}}_{\mathbf{\text{2}}}\mathbf{\text{O(s)}}\mathbf{\to }{\mathbf{\text{6Cu(l) + SO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

(a) Give the oxidation states of copper in${\mathbf{\text{Cu}}}_{\mathbf{\text{2}}}{\mathbf{\text{S, Cu}}}_{\mathbf{\text{2}}}\mathbf{\text{O, and Cu.}}$

(b) What are the oxidizing and reducing agents in this reaction?

Short answer

a) The oxidation states of copper are${\text{+ 1 in Cu}}_{\text{2}}{\text{S and Cu}}_{\text{2}}\text{O and 0 in Cu.}$

b) The ${\text{Cu}}_{\text{2}}\text{S}$is a reducing agent and the ${\text{Cu}}_{\text{2}}\text{O}$is an oxidizing agent.

Step-by-step solution

Concept Introduction

The process in which heat is used to remove a base metal from ore is termed as smelting. Extractive metallurgy is what it is. Metals like silver, iron, copper, and other base metals, can be extracted from their ores usingthe mentioned process.

Copper has two main oxidation states:$\mathbf{\text{+ 1 and + 2,}}$ while rare$\text{+ 3}$ complexes have been discovered.

Obtaining the Oxidation States of Copper in  Cu2S, Cu2O, and Cu

a) Let us find the oxidation states of copper in mentioned compounds,

• ${\text{Cu}}_{\text{2}}\text{S:}$Since oxidation state of $\text{S is - 2,}$the oxidation state of $\text{Cu is + 1}$.
• ${\text{Cu}}_{\text{2}}\text{O}$: Since oxidation state of $\text{O is - 2}$, the oxidation state of $\text{Cu is + 1}$.
• $\text{Cu:}$Since copper is in a liquid, molten state of a pure metal (elemental form)- the oxidation state of $\text{Cu}$is 0.

Finding the Oxidizing and Reducing

b) Let us find the oxidizing and reducing agents in the given reaction,

${\mathrm{Cu}}_2\mathrm{S}\left(\mathrm{s}\right)+2{\mathrm{Cu}}_2\mathrm{O}\left(\mathrm{s}\right)\to 6\mathrm{Cu}\left(\mathrm{l}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)$

The reducing agent is oxidised, which means it is losing electrons. With the addition of 6 electrons, the oxidation state of S changes ${\text{- 2 in Cu}}_{\text{2}}{\text{S to + 4 in SO}}_{\text{2}}$. As a result, we get the reducing agent which is ${\mathrm{Cu}}_2\mathrm{S}$.

As $\mathrm{Cu}$is reduced, the ${\mathrm{Cu}}_2\mathrm{O}$acts as an oxidising agent. With the addition of one electron, the oxidation state of $\mathrm{Cu}$changes from in ${\text{Cu}}_{\text{2}}\text{O to 0 in Cu}$.