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.

Answer

6

Explanation

`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