InterviewSolution
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InterviewSolution
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The emf of a cell corresponding to the reaction `Zn +2H^(+)(aq) rarr Zn^(2+) (0.1M) +H_(2)(g) 1` atm is `0.28` volt at `25^(@)C`. Calculate the `pH` of the solution at the hydrogen electrode. `E_(Zn^(2+)//Zn)^(@) =- 0.76` volt and `E_(H^(+)//H_(2))^(@) = 0`A. 7.05B. 8.62C. 8.75D. 9.57 |
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Answer» Correct Answer - B The half-cell reaction are `Zn to Zn^(2+) + 2e^(-)` `2H^(+) + 2e^(-) to H_2 ` We know that `E_(Zn//Zn^(2+))=E_(Zn//Zn^(2+))^@-"RT"/"nF""In"([Zn^(2+)])/([Zn])` `therefore E_(Zn//Zn^(2+))=0.76-(2.303xx8.314xx298)/(2xx96500)"log"0.1/1` =0.76-(-0.03)=0.79 V Also, `E_(H^(+)//H_2)=E_(H^(+)//H_2)^@-"RT"/"nF""In"([H_2])/([H^+]^2)` `=0-(2.303xx8.314xx298)/(2xx96500)"log"1/([H^+]^2)` =0.0591 log `[H^+]` =-0.0591 pH Since `E_"cell"=E_(Zn//Zn^(2+))+E_(H^+//H_2)` or 0.28=0.79-0.0591 pH or `pH=(0.79-0.28)/0.0591=0.51/0.0591`=8.62 |
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