InterviewSolution
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InterviewSolution
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Consider the following half reactions `:` `PbO_(2)(s)+4H^(o+)(aq)+SO_(4)^(2-)+2e^(-)rarrPbSO_(4)9s)+2H_(2)O" "E^(c-)=+1.70V` `PbSO_(4)(s)+2e^(-)rarr Pb(s)+SO_(4)^(2-)(aq)" "E^(c-)=-0.31V` `a.` Calculate the value of `E^(c-)` for the cell. `b.` Calculate the voltage generated by the cell if `[H^(o+)]=0.10M` and `[SO_(4)^(2-)]=2.0M` `c.` What voltage is generated by the cell when it is at chemical equilibrium ? |
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Answer» `a. E^(c-)._(cell)=E^(c-)._(Cathode)-E^(c-)._(Anode )` `=1.70-(-0.31)=2.01V` `PbO_(2)+4H^(o+)+SO_(4)^(2-)+2e^(-)rarr PbSO_(4)+2H_(2)O` `Pb+SO_(4)^(2-)rarr PbSO_(4)` `ulbar(PbO_(2)+Pb+4H^(o+)+2SO_(4)^(2-)rarr 2PbSO_(4)+2H_(2)O)` `b.` `E_(cell)=E^(c-)._(cell)-(0.059)/(2)log.(1)/([H^(o+)][SO_(4)^(2-)]^(2))` `=2.01-(0.059)/(2)log.(1)/((0.14)^(4)(2.0)^(2))` `c.` When cell is at equilibrium , `E_(cell)=0` |
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