Question
Write two applications of electrochemical series.

Answer

I. Predicting the Liberation of Hydrogen Gas from Acids by Metals : All metals having negative electrode potentials (negative $E _{\text {red }}^0$ ) show greater tendency of losing electrons as compared to hydrogen. So, when such a metal is placed in an acid solution, the metal gets oxidised, and $H ^{+}$ions get reduced to form hydrogen gas. Thus the metals having negative $E _{ red }^0$ values liberate hydrogen from acid solution.
$M \longrightarrow M _{(a q)+n e^{-}}^{n+}$
Metal having negative $E _{\text {red }}^{\circ}$ value
Thus, $\quad n H _{(a q)}^{+}+n e^{-} \longrightarrow n / 2 H _{2(g)}$
$M +n H _{(a q)}^{+} \longrightarrow M _{(a q)}^{n+}+\frac{n}{2} H _{2(g)}$
In other words, we can say that metals which are placed above hydrogen in electro-chemical series liberate hydrogen from acids while the metals which are below hydrogen in the electro-chemical series (like $Cu , Hg , Au$, Pt etc.) do not evolve hydrogen from acids, as they have positive value of electrode reduction potential.
II. Predicting the Displacement of Metals : A metal higher in the series will displace the metal from its solution which is lower in the series i.e., the metal having low standard reduction potential will displace the metal from its salt solution which has higher value of standard reduction potential. This is due to the fact that the metal having more negative value of $E _{\text {red }}^0$ have a higher tendency to lose electrons. Thus, having a greater tendency to dissolve in solution as cations. The ions of another metal having higher $E _{\text {red }}^0$ value show a greater tendency towards reduction.
Example : Zinc displaces copper from $CuSO _4$
Solution
$Zn _{(s)}+ CuSO _{4(a q)} \longrightarrow Cu _{(s)}+ ZnSO _{4(a q)}$
or
Image
The $E _{ red }^0$ of $Zn =-0.76 V$ and for $Cu =+0.34 V$.
Zn has lower $E ^0$ value as compared to $Cu . Zn$ has greater tendency towards oxidation and $Cu ^{2+}$ ions get reduced to Cu metal.
$Zn \longrightarrow Zn ^{2+}+2 e^{-}$ $\quad$(Oxidation)
$Cu ^{2+}+2 e^{-} \longrightarrow Cu$$\quad$(Reduction)
Thus, Low displaces or reduces high
Similarly, copper displaces silver from $AgNO _3$ solution or iron can displace copper from $CuSO _4$ but silver does not displace copper.

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