Question

The metal which displaces hydrogen from a boiling caustic soda solution is
A. As
B. Zn
C. Mg
D. Fe

Answer 1

Hint: In the electrochemical series those elements placed above the hydrogen are more reactive and they have a special ability to displace hydrogen. Here we have four elements in the given options, first, we have to give attention to the electrochemical series to get an idea about their relative positions with respect to hydrogen.

Complete Step by Step Answer:
The electrochemical series describes to us the arrangement of elements in order of their electrode potential values. The elements placed at the top of this reactivity series are good reducing agents and those who are placed below this series are good oxidising agents.

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

From this electrochemical series, it is seen that zinc (Zn) is positioned above hydrogen and has a negative reductive potential value, ${{E}^{O}}=-0.76eV$ . So, it is capable of displacing hydrogen from a boiling caustic soda solution (NaOH) and liberating hydrogen gas.
$Zn(s)+2NaOH(aq.)+2{{H}_{2}}O(aq.)\to N{{a}_{2}}[Zn{{(OH)}_{4}}](s)+{{H}_{2}}(g)$

Though magnesium is positioned above the hydrogen under normal conditions, magnesium metal does not immediately react with boiling caustic soda or sodium hydroxide (NaOH) due to the presence of an oxide layer.

The other two elements Fe and As are not able to displace hydrogen from boiling caustic soda solution as they have lower reactivity than hydrogen and are placed below the electrochemical series.
Thus, option (B) is correct.

Note: Reducing the character of metals depends on the tendency of losing electrons. The higher the negative reductive potential, the higher will be the tendency to lose electrons. So, their reducing character decreases from top to bottom in the electrochemical series. In this way, lithium is the stronger reducing agent and possesses the highest negative ${{E}^{O}}$ .