Consider the half - cell : `Pt|H_(2)"(1 atm)"|H_(2)SO_(4)`. If the reduction potential of the half - cell at `25^(@)C` be 0.3 V, then determine the pH of `H_(2)SO_(4)` solution in the half - cell.

Explanation

The reduction occurring in the half - cell :
`2H^(+)(aq)+2e rarr H_(2)(g)`
The Nernst equation for this reaction is,
`E_("Red")=E_("Red")^(@)-(0.059)/(2)log.(pH_(2))/([H^(+)]^(2))`
For hydrogen electrode, `E_(Red")^(@)=0`
`"or, "0.6=+0.059log[H^(+)]^(2)" or, "log[H^(+)]=-10.169`
`"or, "log[H^(+)]=-5.08" or, "-log[H^(+)]=-5.08 therefore pH=-5.08`