Question

The chemical reaction, \[2{\rm{AgCl}}\left( s \right) + {{\rm{H}}_{\rm{2}}}\left( g \right) \to 2{\rm{HCl}}\left( {aq} \right) + 2{\rm{Ag}}\left( s \right)\] taking place in a galvanic cell is represented by the notation:
A. Pt(s)|\[{{\rm{H}}_{\rm{2}}}\] (g), 1 bar|1 M KCl (aq)|AgCl (s)|Ag (s)
B. Pt(s)|\[{{\rm{H}}_{\rm{2}}}\] (g), 1 bar|1 M HCl (aq)| 1 M \[{\rm{A}}{{\rm{g}}^ + }\] (aq)|Ag(s)
C. Pt(s)|\[{{\rm{H}}_{\rm{2}}}\] (g), 1 bar|1 M HCl (aq)|AgCl (s)|Ag(s)
D. Pt(s)|\[{{\rm{H}}_{\rm{2}}}\] (g), 1 bar|1 M HCl (aq)|AgCl (s)|AgCl (s)

Answer 1

Hint: There are two electrodes namely cathode and anode that consist of an electrochemical cell. These electrodes are dipped in a solution called electrolytes and two half cells are formed. When these two half cells are combined it leads to the formation of a cell.

Complete Step by Step Solution:
Let's discuss the rules of representation of cells.
1) In a galvanic cell representation, anode, where oxidation occurs, is to be written on the left and cathode is to be written on the right-hand side.
2) The representation of the anode is done by writing the solid phase of the metal and then the cation of the metal. And the representation of the cathode is done by first writing the cation and then the metal.
3) The separation of metal and the cation is done by using a vertical line.
4) The salt bridge that does the function of the separation of two half cells is represented by two vertical lines.

In the given question, the reaction is \[2{\rm{AgCl}}\left( s \right) + {{\rm{H}}_{\rm{2}}}\left( g \right) \to 2{\rm{HCl}}\left( {aq} \right) + 2{\rm{Ag}}\left( s \right)\]
Here, reduction of Silver takes place, that is, \[{\rm{A}}{{\rm{g}}^ + } \to {\rm{Ag}}\] , so, it is to be written in right hand side of the cell notation. So, the silver electrode acts as a cathode.
And the platinum electrode undergoes oxidation, so it will be written on the left hand side.
Therefore, the correct cell notation is Pt(s)|\[{{\rm{H}}_{\rm{2}}}\] (g), 1 bar|1 M HCl (aq)| 1 M \[{\rm{A}}{{\rm{g}}^ + }\] (aq)|Ag(s).
Hence, option B is right.

Note: In the electrochemical cells, platinum is used because of its nature to resist oxidation. It does not undergo chemical reactions easily. So, it will not become part of the redox reactions occurring in the electrochemical cells.