Question

Hydrogen cannot reduce
A. Hot CuO
B. $F{{e}_{2}}{{O}_{3}}$
C. Hot $Sn{{O}_{2}}$
D. Hot $A{{l}_{2}}{{O}_{3}}$

Answer 1

Hint: Hydrogen can not reduce those elements which have lower reduction potential value than it in the Electrochemical series. To solve the above question, first, we have to give focus on the electrochemical series. It helps us to identify the substance that has a greater reduction potential value than hydrogen.

Complete Step by Step Answer:
Electrochemical series helps us to identify the substances which are good reducing agents and those which are good oxidising agents. In this reactivity series, substances placed below the hydrogen are easy to be reduced but those above the hydrogen are very much difficult to reduce.

Electrodes	${{E}^{O}}(Volts)$
$L{{i}^{+}}/Li$	$-3.04$
${{K}^{+}}/K$	$$$-2.92$
$C{{a}^{2+}}/Ca$	$-2.87$
$M{{g}^{2+}}/Mg$	$-2.37$
$A{{l}^{3+}}/Al$	$-1.66$
$Z{{n}^{2+}}/Zn$	$-0.76$
${{H}^{+}}/H$	$0.00$
$S{{n}^{4+}}/S{{n}^{2+}}$	$+0.14$
$C{{u}^{2+}}/Cu$	$+0.34$
$F{{e}^{3+}}/F{{e}^{2+}}$	$+0.77$
$A{{g}^{+}}/Ag$	$+0.80$
$P{{t}^{2+}}/Pt$	$+1.20$
${{F}_{2}}/{{F}^{-}}$	$+2.87$

Fig 1: Electrochemical series

In this electrochemical series, we see that aluminium (Al) is placed above the hydrogen, possessing electrode reduction potential $-1.66eV$ . Other substances like iron (Fe), tin (Sn), and copper(Cu) have reduced potential values are $+0.77,+0.14$and $+0.34eV$respectively. We also know the fact that those who have the most negative or less reduction potential value tend to be a better reducing agent than others.

So, here hydrogen can not reduce $A{{l}^{3+}}$ in a hot $A{{l}_{2}}{{O}_{3}}$ solution, as Al has a negative reduction potential than hydrogen (zero reduction potential value) But it can reduce the other three substances such as $C{{u}^{2+}}$ to Cu from hot CuO solutions, $S{{n}^{4+}}$ to $S{{n}^{2+}}$ from hot $Sn{{O}_{2}}$ solutions, and $F{{e}^{3+}}$ to $F{{e}^{2+}}$ form the solution of $F{{e}_{2}}{{O}_{3}}$.
Thus, option (D) is correct.

Note: It is not mandatory to remember the exact values of substances in electrochemical series but having a general idea of the relative positions of various elements in this reactivity series will help us in determining the products of redox reactions i.e we will know beforehand which reactant is going to get reduced and which will get oxidised.