The e.m.f of a cell corresponding to the reaction
`Zn+2H^(+)(aq)toZn^(2+)(0.1M)+H_(2)(g)`(1 atm)
is 0.26 volt at `25^(@)C`.Calculate the pH of the solution at the hydrogen electrode.

Answer

9

Explanation

`E_(cell)^@=0.76` volt
Applying nernst equation
`E_(cell)=E_(cell)^@(-0.0591)/2"log"([Zn^(2+)][P_(H_2)])/([H^+]^2), " " 0.26=0.76(-0.0591)/2"log"((0.1)(1))/([H^+]^2)`
log `0.1/([H^+]^2)=(2xx0.50)0.0591`
or log 0.1-log `[H^+]^2=17` or 2pH=17-log 0.1
pH=`18/2=9`