A) \[\Pr (s)|{{H}_{2}}(g),1bar|1\,M\,KCl(aq)|\] \[AgCl(s)|Ag(s)\]
B) \[\Pr (s)|{{H}_{2}}(g),\,\,\,1bar|1\,M\,HCl(aq)|\] \[1M\,A{{g}^{+}}(aq)|Ag(s)\]
C) \[\operatorname{Pt}(s)|{{H}_{2}}(g),\,1bar|1M\,HCl(aq)|\] \[AgCl(s)|Ag(s)\]
D) \[\Pr (s)|{{H}_{2}}(g),1bar|1\,M\,HCl(aq)]\] \[Ag(s)|AgCl(s)\]
Correct Answer: B
Solution :
\[2AgCl(s)+{{H}_{2}}(g)\,(1\,bar)\xrightarrow{{}}\] \[2HCl(aq)+2\,Ag(s)\] The cell with this cell reaction can be represented as: \[\underset{(anode)}{\mathop{Pt(s)}}\,|{{H}_{2}}(g)\,(1\,Bar)|1\,MHCl(aq),\] \[\underset{(cathode)}{\mathop{||A{{g}^{+}}(aq)||}}\,Ag(s)\] Silver is undergoing reduction \[(A{{g}^{+}}\xrightarrow{{}}Ag)\] 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 \[AgCl\] (5) and \[AgCl\] do not ionise, it is also incorrect, (d) at cathode Ag is being oxidized to \[A{{g}^{+}}\] and at cathode oxidation does not take place hence it is also incorrect. Note: Remember \[LEO\xrightarrow{{}}\]loss of electrons is oxidation.
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