The emf of the cell : `H_(2)(g)|"Buffer||Normal"` canlomel elctrolde is `0.6885V` at `25^(@)C`, when barometric pressure is `=760 mm`. What is the pH of the buffer solution? `E_("calomel")=0.28V`

Options

A. `9.92`
B. `7.92`
C. `8.92`
D. `6.92`

Answer

D

Explanation

As the right electrode is normal (standard) electrode, the emf of cell may be written as
`E_(cell)=E_("calomel")-E_(H^(+)("buffer")//H_(2))E_(cell)=E_("calomel")-E_(H^(+)("buffer")//H_(2))`
or, `0.6885=0.28-E_(H^(+)//H_(2))`
`thereforeE_(H^(+)//H_(2))=-0.4085V`
Now, from Nernst equation,
`E_(H^(+)//H_(2))=E_(H^(+)//H_(2))^(0)-(0.059)/(n)log.(P_(H_(2)))/([H^(+)]^(2))`
`-0.4085=0-(0.059)/(2)`
`or,(logP_(H_(2))-log[H^(+)]^(2))`
or, `0.4085=(0.059)/(2)(log1+2P^(H))`
`therefore`pH of buffer solution = `6.92`