`Zn(s)+2H^(+)(aq)rarr Zn^(2+)(0.1M)+H_(2)(g,"1 atm")`
Cell potential of a chemical cell in which the above reaction occurs is `+0.28V` at `25^(@)C`. Write the half-cell reactions and calculate the pH of the hydrogen electrode.
Give : `E_("Zn^(2+)||Zn)^(@)=-0.76V and E_(2H^(+)||H)^(@)=0.00V`

Explanation

`Zn(s)+2H^(+)(aq)rarr Zn^(2+)(0.1M)+H_(2)("g, 1 atm")`
Oxidation half - cell reaction:
`Zn(s)rarr Zn^(2+)(0.1M)+2e`
Reduction half cell reaction:
`2H^(+)(aq)+2e rarr H_(2)("g, 1 atm")`
`therefore" "E_("cell")^(@)=E_(2H^(+)//H_(2))^(@)-E_(Zn^(2+)//Zn)^(@)`
`=0.00V-(-0.76V)=0.76V`
As given, `E_("cell")=0.28V`
Therefore, at `25^(@)C`, according to Nernst equation,
`E_("cell")=E_("cell")^(@)-(0.059)/(2)log.([Zn^(2+)](p_(H_(2))))/([Zn][H^(+)]^(2))`
[The redox reaction occurs by transfering 2 electrons]
`"or, "0.28=0.76-(0.059)/(2)log.(0.1xx1)/(1xx[H^(+)]^(2))`
`" "[because " concentration of solid = 1 "]`
`"or, "(0.059)/(2)log.(1)/(10xx[H^(+)]^(2))=0.76-0.28`
`"or, "-log(10xx[H^(+)]^(2))=(0.48xx2)/(0.059)=16.27`
`"or, "-log10-2log[H^(+)]=16.27`
`"or, "-1-2log[H^(+)]=16.27`
`"or, "-2log[H^(+)]=17.27`
`"or, "-log[H^(+)]=8.635 or, pH=8.635`