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Given: E(Sn^(2+)//Sn) ^0 = -0.14V , E(Pb...

Given: `E_(Sn^(2+)//Sn) ^0 = -0.14V` , `E_(Pb^(2+)//Pb) ^0 = -0.13V`. Determine `[(Sn^(2+))//(Pb^(2+))]` at equilibrium. For cell reaction `Sn |Sn^(2+) || Pb^(2+) | Pb` , Take `(2.303RT)//F = 0.06V`

Text Solution

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For the given cell, `E_(cell)^(@)=E_(Pb^(2+)//Pb)^(@)-E_(Sn^(2+)//Sn)^(@)=0.126-(-0.136)=0.01V`
For the given cell reaction,
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Sn^(2+)])/([Pb^(2+)])=0.01-(0.0591)/(2)"log"([Sn^(2+)])/([Pb^(2+)])=0.01+0.0295"log"([Pb^(2+)])/([Sn^(2+)])`
At equilibrium `E_(cell)=0`
`therefore0.01+0.0295"log"([Pb^(2+)])/([Sn^(2+)])=0` or `"log"([Pb^(2+)])/([Sn^(2+)])=-(0.01)/(0.0295)=-0.3390=overline(1).6610`
`therefore([Pb^(2+)])/([Sn^(2+)])="Antilog "overline(1).6610=0.458`
Thus, so long as `([Pb^(2+)])/([Sn^(2+)]) gt0.458`, the cell reaction as given will place. when `([Pb^(2+)])/([Sn^(2+)]) lt 0.458,E_(cell)` will become -ve, i.e., the reaction will be reversed.
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