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.

Explanation

Anode reaction `:` `(1)/(2)H_(2)(g)rarr H^(o+)+cancel (e^(-))`
Cathode reaction
`AgCl(s)+cancel(e^(-))rarrAg(s)+Cl^(c-)`
Cell reaction `:`
`ulbar((1)/(2)H_(2)(g)+AgCl(s) rarr Ag(s)+Cl^(c-)+H^(o+))`
`E^(c-)._(cell)+(E^(c-)._(reduction ))_(c)-(E^(c-)._(reduction ))_(a)`
`=0.23-0=0.23V`
`E_(cell)=E^(c-)._(cell)-(0.059)/(n_(cell))log [H^(o+)][Cl^(c-)]`
`[n_(cell)=1,` activity of `Ag(s) ` and `AgCl(s)=1,p_(H_(2))=1atm]`
`1.05V=0.23-(0.059)/(1)log[H^(o+)][Cl^(c-)]`
`[{: (K_(w)=[H^(o+)][overset(c-)(O)H]),( :. [H^(o+)][Cl^(c-)]=(K_(W)[Cl^(c-)])/([overset(c-)(O)H])):}]`
`1.05V=0.23-0.059log .(K_(W)[Cl^(c-)])/([overset(c-)(O)H])` `[{: ([Cl^(c-)]=[overset(c-)(O)H]),( =0.1M):}]`
`1.05V=0.23-0.059log .(K_(W)xx0.1M)/(0.1M)`
`(1.05-0.23)V=0.059[-log K_(W)]=0.059pK_(W)`
`:. pK_(W)=((1.05-0.23))/(0.059)=13.89`