InterviewSolution
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InterviewSolution
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Calculate the cell potential of half cell having at the reaction : `M_(2)S+2e^(-)rarr2M+S^(2-)` at `27^(@)C` in a solution of pH = 3 and saturated with `0.1 MH_(2)S` for `H_(2)SK_(1)=10^(-8)` and `K_(2)=10^(-13)Ksp (M_(2)S)=1.0 ,10^(-50)E_(M^(+)//M)^(@)=1V` assume R = 10 J/K/mol and `(2.303RT)/(nF)=(0.07)/(n)` Express the magnitued of your answer after multiplication with 25 and as nearest highest whole number. |
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Answer» Correct Answer - 6 `M_(2)S hArr 2M^(+)+S^(2)Delta G_(1)^(@)` `{:(" "2M^(+)+2e^(-)rarr2M Delta G_(2)^(@)),(bar(M_(2)S+2e^(-)rarr2M+S^(2-)Delta G_(3)^(@))):}` `Delta G_(3)^(@)=Delta G_(1)^(@)+Delta G_(3)^(@)` `Delta G_(1)^(@)=-2.303xx10xx300 log 10^(-50)` `=2.303xx3000xx50` `=34.545xx10^(4)J` `Delta G_(2)^(@)=-2xx96.500xx1=-193000` `Delta G_(3)^(@)=-19300+345450` `Delta G_(3)^(@)=152450-nFE_(3)^(@)=152450` `E_(3)^(@)=(-152450)/(2xx96,500)=-(15245)/(19300)` `=-0.79` `~~-0.8` `E_("cell")=E_("cell")^(@)-(0.07)/(n)log [S^(2-)]` `=-0.8-(0.07)/(2)log 0.2` `K_(1)K_(2)=([H^(+)]^(2)[S^(2-)])/([H_(2)S])` `=(10^(-6)[S^(2-)])/(0.1)=[S^(2-)]=10^(-16)` `E_("cell")=-0.8-(0.07)/(2)log 10^(-16)` `=-0.8+(0.07)/(2)xx-16` `=-0.8+0.07xx8 = -0.24V` After multiplication with 25, the magnitude of above answer becomes 6 |
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