Question

The chemical reaction,
$2AgC{l}_{(s)}+H_{2(g)}\to 2HC{l}_{(aq)}+2A{g}_{(s)}$
taking place in a galvanic cell is represented by the notation

• A. $P{t}_{(s)}|H_{2(g)},1bar|1MKC{l}_{(aq)}|AgC{l}_{(s)}|A{g}_{(s)}$
• B. $P{t}_{(s)}|H_{2(g)},1bar|1MHC{l}_{(aq)}|1MA{g}_{(aq)}^{+}|A{g}_{(s)}$
• C. $P{t}_{(s)}|H_{2(g)},1bar|1MHC{l}_{(aq)}|AgC{l}_{(s)}|A{g}_{(s)}$
• D. $P{t}_{(s)}|H_{2(g)},1bar|1MHC{l}_{(aq)}|A{g}_{(s)}|AgC{l}_{(s)}$

Answer 1

Answer: (B)

$2AgCl(s)+H_2(g)(1bar)⟶2HCl(aq)+2Ag(s)$
The cell with this cell reaction can be represented as :
$\underset{\text{(anode)}}{Pt(s)}\left|H_2(g)(1Bar)\right|1MHCl(aq)$,
$\Vert \underset{\text{ (cathode) }}{\Vert A{g}^{+}(aq)\Vert Ag(s)}$
silver is undergoing reduction $(A{g}^{+}⟶Ag)$
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 $A{g}^{+}$ and at cathode oxidation does not take place hence it is also incorrect.
Remember $LEO⟶$ loss of electrons is oxidation.