InterviewSolution
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InterviewSolution
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The EMF of a cell corresponding to the reaction, `Zn(s)+2H^(+)(aq)toZn^(2+)(0.1M)+H_(2)(g,1atm)` is 0.28 volt at `25^(@)C`. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode. `E_(Zn^(2+)//Zn)^(@)=-0.76"volt",E_(H^(+)//H_(2))^(@)=0` |
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Answer» The half cell reaction are: `ZntoZn^(2+)+e^(-)` (Oxidation half reaction) `2H^(+)+2e^(-)toH_(2)` (Reduction half reaction) The cell may be represented as: `Zn|Zn^(2+)||H^(+)|H_(2)` For the given cell reaction, `E_(cell)^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-E_(Zn^(2+)//Zn)^(@)=0-(-0.76V)=0.76V` Applying Nernst eqn. to the given cell reaction, `E_(cell)=E_(cell)^(@)=(0.0591)/(2)"log"([Zn^(2+)])/([H^(+)]^(2))` `therefore0.28=0.76-(0.0591)/(2)"log"((0.1))/([H^(+)]^(2))=0.76-(0.0591)/(2)[log0.1-2log(H^(+))]` `=0.76-0.02955(-1+2pH)` `[becausepH=-log[H^(+)]]` `=0.76+0.2095-0.0591pH` or `pH=(0.5095)?(0.591)=8.62` |
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