Question

The chemical reaction,
$2AgCl_{(s)} + H_{2(g)} \to 2HCl_{(aq)} + 2Ag_{ (s)}$
taking place in a galvanic cell is represented by the notation

• A. $Pt_{(s)} | H_{2(g)}, 1 bar | 1M KCl_{(aq)} | AgCl_{(s)} | Ag_{(s)}$
• B. $Pt_{(s)} | H_{2(g)}, 1 bar | 1M HCl _{(aq)} | 1M Ag^+_{ (aq)} | Ag_{(s)}$
• C. $Pt_{(s)} | H_{2(g)}, 1 bar | 1M HCl _{(aq)} | AgCl_{(s)} | Ag_{(s)}$
• D. $Pt_{(s)} | H_{2(g)}, 1 bar | 1M HCl _{(aq)} | Ag_{(s)} | AgCl_{(s)}$

Answer 1

Answer: (B)

$2 AgCl (s)+ H _{2}(g)(1 bar ) \longrightarrow 2 HCl (a q)+2 Ag (s)$
The cell with this cell reaction can be represented as :
$\underset{\text{(anode)}}{Pt(s)}\left| H _{2}(g)(1 Bar )\right| 1 MHCl (a q)$,
$\| \underset{\text { (cathode) }}{\left\| Ag ^{+}(a q)\right\| Ag (s)}$
silver is undergoing reduction $(Ag ^{+} \longrightarrow Ag)$
in this reaction, hence it will act as cathode in
the cell.
Option (a) has $KCl$ which is not present in the cell, so it is incorrect. (c) has $AgCl (s)$ and $AgCl$ do not ionise, it is also incorrect. (d) at cathode $Ag$ is being oxidised to $Ag ^{+}$ and at cathode oxidation does not take place hence it is also incorrect.
Remember $LEO \longrightarrow$ loss of electrons is oxidation.