The overall reaction electrochemical cell at 298K. `Ag(s)|AgI(s)|I^(-)(aq)||Cl^(-)(aq)|Hg_(2)Cl_(2)|Hg(l)|Pt(s)` [Given: `E_(Cl^(-)|Hg_(2)Cl_(2)|Hg)^(

Question

The overall reaction electrochemical cell at 298K. `Ag(s)|AgI(s)|I^(-)(aq)||Cl^(-)(aq)|Hg_(2)Cl_(2)|Hg(l)|Pt(s)` [Given: `E_(Cl^(-)|Hg_(2)Cl_(2)|Hg)^(

The overall reaction electrochemical cell at 298K.
`Ag(s)|AgI(s)|I^(-)(aq)||Cl^(-)(aq)|Hg_(2)Cl_(2)|Hg(l)|Pt(s)`
[Given: `E_(Cl^(-)|Hg_(2)Cl_(2)|Hg)^(@)=0.26V. E_(Ag^(+)|Ag)^(@)=0.8V`.
`K_(sp)(Agl)=10^(-16)` and `(2.303RT)/(F)=0.06`]
At equilibrium ratio of `([Cl^(-)])/([I^(-)])` in the above cell will be:

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

(i) `Q=([Hg_(2)^(2+)])/([Ag^(+)]^(2))=10^(-1)/([10^(-4)]^(2))=10^(7)`
`E^(@)=E_(Ag^(+)//Ag)^(@)-E_(Hg_(2)^(2+)//Hg)^(@)`
`=0.80-0.79=0.01 V`
`E=E^(@)-0.059/n log Q`
`=0.01-0.059/2 log 10^(7)`
`=-0.1965 V`
Negative value shows that the reaction will proceed from right to left, i.e., in backward direction.
(ii) `Q=([Hg_(2)^(2+)])/([Ag^(+)]^(2))=10^(-4)/([10^(-1)]^(2))=10^(-2)`
`n=2`
`E^(@)=0.01" volt"`
`E=E^(@)-0.059/n log_(10) Q`
`=0.01-0.059/2 log_(10) 10^(-2)`
`=0.01 + 0.059 V`
`=0.069 V`
Since, the value of cell potential is positive, the reaction will proceed spontaneously in forward direction.