ElectrochemistryEasy
Question
The chemical reaction,
2AgCl(s) + H2(g) → 2HCl(aq) = 2Ag(s)
Taking place in a galvanic cell is represented by the notation :
2AgCl(s) + H2(g) → 2HCl(aq) = 2Ag(s)
Taking place in a galvanic cell is represented by the notation :
Options
A.Pt(s)|H2(g)′ 1 bar|1MKCl(aq)|AgCl(s)|Ag(g)
B.Pt(s)|H2(g)′ 1 bar|1MHCl(aq)|1MAg+(aq)|Ag(g)
C.Pt(s)|H2(g)′ 1 bar|1MHCl(aq)|AgCl(s)|Ag(g)
D.Pt(s)|H2(g)′ 1 bar|1MHCl(aq)|Ag(g)|AgCl(s)
Solution
In the given reaction 2AgCl(s) + H2(g)(1Bar) 
2HCl(aq) + 2Ag(s)
Silver is undergoing reduction(Ag+ → Ag)
hence it will act as cathode in the following cell.
Pt(s)|H2(g)1 bar|1MHCl(aq)|1MAg+(aq)|Ag(g)
(anode) (cathode)
(i) Thus, since option (a) has KCl which is not present in the cell, it is incrrect.
(ii) Since option (c) has AgCl(s) which does not ionise, it is incorrect,
(iii)Since in option(d),at cathode, Ag is being oxidised to Ag+ which is not possitive, so it is also incorrect.
2HCl(aq) + 2Ag(s) Silver is undergoing reduction(Ag+ → Ag)
hence it will act as cathode in the following cell.
Pt(s)|H2(g)1 bar|1MHCl(aq)|1MAg+(aq)|Ag(g)
(anode) (cathode)
(i) Thus, since option (a) has KCl which is not present in the cell, it is incrrect.
(ii) Since option (c) has AgCl(s) which does not ionise, it is incorrect,
(iii)Since in option(d),at cathode, Ag is being oxidised to Ag+ which is not possitive, so it is also incorrect.
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