Question

Which of the following is a feasible reaction?
(A) \[{\text{Ba(s) + }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{(aq)}} \to {\text{BaS}}{{\text{O}}_{\text{4}}}{\text{(aq) + 2K(s)}}\]
(B) \[{\text{Zn(s) + 2AgN}}{{\text{O}}_{\text{3}}}{\text{(aq)}} \to {\text{Zn(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{(aq) + 2Ag(s)}}\]
(C) \[{\text{Mg(s) + N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{(aq)}} \to {\text{MgS}}{{\text{O}}_{\text{4}}}{\text{(aq) + 2Na(s)}}\]
(D) \[{\text{Cu(s) + MgS}}{{\text{O}}_{\text{4}}}{\text{(aq)}} \to {\text{CuS}}{{\text{O}}_{\text{4}}}{\text{(aq) + Mg(s)}}\]

Answer 1

Hint: A metal can displace the other metal from its salt solution, provided, the other metal is placed above the first metal in the electrochemical series. This is because the cell potential will be positive and the standard Gibbs free energy change will be negative.

Complete answer:
Consider the reaction between barium metal and potassium sulphate to form barium sulphate and potassium metal.
\[{\text{Ba(s) + }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{(aq)}} \to {\text{BaS}}{{\text{O}}_{\text{4}}}{\text{(aq) + 2K(s)}}\]
Barium metal cannot displace potassium metal from potassium sulphate, as potassium metal is placed below the barium metal in the electrochemical series.
Consider the reaction between zinc metal and silver nitrate to form zinc nitrate and silver metal.
\[{\text{Zn(s) + 2AgN}}{{\text{O}}_{\text{3}}}{\text{(aq)}} \to {\text{Zn(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{(aq) + 2Ag(s)}}\]
Zinc metal can displace silver metal from silver nitrate, as silver metal is placed above the zinc metal in the electrochemical series.
Consider the reaction between magnesium metal and sodium sulphate to form magnesium sulphate and sodium metal.
\[{\text{Mg(s) + N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{(aq)}} \to {\text{MgS}}{{\text{O}}_{\text{4}}}{\text{(aq) + 2Na(s)}}\]
Magnesium metal cannot displace sodium metal from sodium sulphate, as sodium metal is placed below the magnesium metal in the electrochemical series.
Consider the reaction between copper metal and magnesium sulphate to form copper sulphate and magnesium metal.
\[{\text{Cu(s) + MgS}}{{\text{O}}_{\text{4}}}{\text{(aq)}} \to {\text{CuS}}{{\text{O}}_{\text{4}}}{\text{(aq) + Mg(s)}}\]
Copper metal cannot displace magnesium metal from magnesium sulphate, as magnesium metal is placed below the copper metal in the electrochemical series.
Thus, the following is a feasible reaction.
\[{\text{Zn(s) + 2AgN}}{{\text{O}}_{\text{3}}}{\text{(aq)}} \to {\text{Zn(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{(aq) + 2Ag(s)}}\]

Hence, the correct option is the option (B).

Note: When the second metal is displaced from its salt solution, it is reduced. This is because the ion of the second metal accepts one or more electrons. So first metal acts as a reducing agent. The metals placed at the bottom of the electrochemical series are strong reducing agents as they are highly electropositive alkali metals.