Question

Which metals cannot be used as a reducing agent in smelting?
A. C
B. Al
C. Cu
D. Na

Answer 1

Hint: Smelting is a type of extractive metallurgical process, which involves the extraction of base metal from ore. Here reducing agent plays an important role to decompose the ore and eliminating other elements such as slag or gases produces a pure base metal. Reducing agents are those chemical species that donate an electron to an electron recipient.

Complete Step by Step Answer:
A metallurgical process of smelting involves the extraction of base metal from ore. This process uses a chemical reducing agent and heat to change the oxidation number of the metal ore. Reduction is the last step of the smelting process. It is the step when oxide becomes the elemental metal. Such an environment pulls the metal from its oxide ore. Most of the time carbon monoxide is used as a reducing agent.
The reducing character of any element can be explained by electrochemical series. Strong reducing agents have negative reduction potential and they are placed below the hydrogen in the electrochemical series.

Electrodes	Standard potential value,${{E}^{{\mathrm O}}}$($V$)
${{F}_{2}}/{{F}^{-}}$	$+2.87$
$C{{l}_{2}}/C{{l}^{-}}$	$+1.36$
$A{{g}^{+}}/Ag$	$+0.80$
$C{{u}^{2+}}/Cu$	$+0.33$
$2{{H}^{+}}/{{H}_{2}}$	$0.00$
$P{{b}^{2+}}/Pb$	$-0.13$
$F{{e}^{2+}}/Fe$	$-0.44$
$Z{{n}^{2+}}/Zn$	$-0.76$
$A{{l}^{3+}}/Al$	$-1.66$
$M{{g}^{2+}}/Mg$	$-2.36$
$L{{i}^{+}}/Li$	$-3.05$

Figure: Electrochemical series

Here from the table, we can see that elements such as lithium ($Li$), sodium ($Na$), magnesium ($Mg$), aluminium (Al), zinc$(Zn)$, etc. are placed below the hydrogen and have negative electrode potential value. That’s why they are good reducing agents.

 Whereas elements like Fluorine ($F$), chlorine ($Cl$), oxygen, copper ($Cu$), etc are placed at the top of hydrogen and have positive electrode potential value. So these elements can not act as a reducing agent, they are oxidising agents.

Therefore sodium ($Na$) and aluminium ($Al$) are reducing agents. Carbon ($C$) which is a non-metal is also a good reducing agent at a higher temperature. It combines with an oxygen $(O)$atom and produces carbon dioxide. But only $(Cu)$ is not a reducing agent having a positive electrode potential value, ${{E}^{O}}=+0.34V$.
Thus, option (C) is correct.

Note: In the Electrochemical series also helps us in determining the feasibility of the reaction. whether the reactants will react to form a product or need some energy to cause the reaction to occur by the equation $\Delta G=nF{{E}^{O}}$ where ${{E}^{O}}$denotes the standard reduction potential of redox reaction.