Question

In which of the following cells, use of salt bridge to eliminate liquid junction potential is not necessary?
A) \[{\text{Pt}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)\parallel {\text{HCl}}\parallel {\text{AgCl}}\left( {\text{s}} \right)\parallel {\text{Ag}}\]
B) \[{\text{Hg/H}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{s}} \right){\text{/KClsaturated/HCl}}\left( {{\text{aq}}} \right){\text{/}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right){\text{Pt}}\]
C) \[{\text{Zn/ZnS}}{{\text{O}}_{\text{4}}}\parallel {\text{CuS}}{{\text{O}}_{\text{4}}}{\text{/Cu}}\]
D) \[{\text{Zn/ZnS}}{{\text{O}}_{\text{4}}}{\text{/ F}}{{\text{e}}^{{\text{3 + }}}}\left( {{\text{aq}}} \right){\text{,F}}{{\text{e}}^{{\text{2 + }}}}\left( {{\text{aq}}} \right){\text{ / Pt}}\]

Answer 1

Hint: We need to know that the redox reaction is one of the type of the type of the type of chemical reaction. In this reaction reduction followed by oxidation in the reactants. The oxidation means the addition of oxygen or the removal of hydrogen or loss of electrons. The reduction means the addition of hydrogen or the removal of oxygen or the gain of electrons. The abbreviation of emf is electromotive force.

Complete answer:
The given cells are
\[{\text{Hg/H}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{s}} \right){\text{/KClsaturated/HCl}}\left( {{\text{aq}}} \right){\text{/}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right){\text{Pt}}\]
\[{\text{Zn/ZnS}}{{\text{O}}_{\text{4}}}\parallel {\text{CuS}}{{\text{O}}_{\text{4}}}{\text{/Cu}}\]
 \[{\text{Zn/ZnS}}{{\text{O}}_{\text{4}}}{\text{/F}}{{\text{e}}^{{\text{3 + }}}}\left( {{\text{aq}}} \right){\text{,F}}{{\text{e}}^{{\text{2 + }}}}\left( {{\text{aq}}} \right){\text{/Pt}}\]
\[{\text{Pt}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)\parallel {\text{HCl}}\parallel {\text{AgCl}}\left( {\text{s}} \right)\parallel {\text{Ag}}\].
The cathode and anode are important for the cell of electrolyte. The cathode is a negatively charged electrode in the cell. The anode is a positive charged electrode in the cell.
The cathode (negative charged electrode) in the cell attaches the positive ions in the electrolytic solution. The anode (positive charged electrode) in the cell attaches the negative ions in the electrolytic solution.
The anion is a negative ion in the electrolytic solution of the cell. The cation is positive ion in the electrolytic solution of the cell
The cells use of salt bridge to eliminate liquid junction potential is necessary given below, \[{\text{Hg/H}}{{\text{g}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{s}} \right){\text{/KClsaturated/HCl}}\left( {{\text{aq}}} \right){\text{/}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right){\text{Pt}}\], \[{\text{Zn/ZnS}}{{\text{O}}_{\text{4}}}\parallel {\text{CuS}}{{\text{O}}_{\text{4}}}{\text{/Cu}}\]and \[{\text{Zn/ZnS}}{{\text{O}}_{\text{4}}}{\text{/F}}{{\text{e}}^{{\text{3 + }}}}\left( {{\text{aq}}} \right){\text{,F}}{{\text{e}}^{{\text{2 + }}}}\left( {{\text{aq}}} \right){\text{/Pt}}\]
\[{\text{Pt}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)\parallel {\text{HCl}}\parallel {\text{AgCl}}\left( {\text{s}} \right)\parallel {\text{Ag}}\] is the one of the given cell use of salt bridge to eliminate liquid junction potential is not necessary.
According to the above discussion, we conclude the cell use of salt bridge to eliminate liquid junction potential is not necessary is \[{\text{Pt}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)\parallel {\text{HCl}}\parallel {\text{AgCl}}\left( {\text{s}} \right)\parallel {\text{Ag}}\].

Hence option A is the correct answer.

Note:
We must have to know that acid reacts with base means to give salt and water. According to this discussion, when Sodium hydroxide reacts with hydrochloric acid to produce a product of sodium chloride as salt and water. Salt reacts with water to give acid and base. According to the second discussion, sodium chloride as salt reacts with water to produce the product of Sodium hydroxide and hydrochloric acid. It is a reversible reaction.
\[{\text{HCl + NaOH}} \rightleftarrows {\text{NaCl + }}{{\text{H}}_{\text{2}}}{\text{O}}\]