The `EMF` of the following cellis `1.05V` at `25^(@)C:`
`Pt,H_(2)(g)(1.0 atm)|NaOH(0.1m),NaCl(0.1M)|AgCl(s),Ag(s)`
`a.` Write the cell reaction,
`b.` Calculate `pK_(w)` of water.

Answer

`pH=8.63`

Explanation

Half cell reactions are `:`
`[Zn(s) rarr Zn^(2+)(aq)+2e^(-)]` at anode
`[H^(o+)+e^(-)rarr (1)/(2)H_(2)]` at cathode
For zinc electrode `:`
`E_(Zn\Zn^(2+))=E^(c-)._(Zn|Zn^(2+))-(0.0591)/(n)log[(Zn^(2+))/(Zn(s))]`
`=0.76-(0.0591)/(2)log .(0.1)/(1)`
`=0.76-(0.0591)/(2)xx(-1)`
`=+0.76+0.03=0.79V`
Similarly, for hydrogen electrode
`E_(H^(o+)|H_(2))=0-(0.0591)/(2)log .(1)/([H^(o+)]^(2))`
or `E_(H^(o+)|H_(2))=0-(0.0591)/(2)xx(-log[H^(o+)])`
`:. E_(H^(o+)|H_(2))=-0.0591pH`
Now, `,E_(cell)=E_(Zn|Zn^(2+))-E_(H^(o+)|H_(2))`
`:.0.28=+0.79-0.0591pH`
or `0.0591pH=+0.79-0.28`
or ` 0.0591pH=0.51`
`:.pH=(0.51)/(0.0591)=8.63`