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For the cell reaction, Sn(s)+Pb^(2+)(aq)toSn^(2+)(aq)+Pb(s), E_(Sn^(2+)//Sn)^(@)=-0.136V, E_(Pb^(2+)//Pb)^(@)=-0.126V, Calculate the ratio of concentration of Pb^(2+) to Sn^(2+) ion at which the cell reaction will be reversed? |
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Answer» SOLUTION :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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