In a cell reduction occurs at the cathode. In this case this will be the hydrogen/Pt half cell. We can find the #sf(H^+)# concentration and hence the pH.
At #sf(25^@C)# The Nernst Equation can be written:
#sf(E_(cell)=E_(cell)^(@)-0.05916/(z)logQ)#
#sf(z)# is the no. of moles of electrons transferred
#sf(Q)# is the reaction quotient
The cell reaction when it is working is:
#sf(Zn+2H^(+)rarrZn^(2+)+H_2)#
We know it proceeds in this direction since #sf(E_(cell))# is +ve.
The reaction quotient is given by:
#sf(Q=([Zn^(2+)]xxp_(H_2))/([H^+]^(2)))#
We can use partial pressures and concentrations together in this expression as they are normalised against the standard conditions of 1 atmosphere and unit concentration.
This becomes:
#sf(Q=1/[H^+]^2)#
So The Nernst Equation becomes:
#sf(E_(cell)=E_(cell)^(@)-0.05916/(z)log[1/[H^+]^2]#
The standard electrode potentials are:
#sf(" "E^@""(V))#
#sf(Zn^(2+)+2erightleftharpoonsZn" "-0.76)#
#sf(2H^(+)+2erightleftharpoonsH_2" "0.00")#
To find #sf(E_(cell)^@# you subtract the least positive potential from the most positive #sf(rArr)#
#sf(E_(cell)^@=0.00-(-0.76)=+0.76color(white)(x)V)#
We can now put the numbers in to solve for #sf([H^+])# and hence find the pH.
#sf(0.58=0.76-0.05916/(2)log[1/[H^+]^2]#
#sf(-0.18=-0.02958log[1/[H^+]^2])#
#:.##sf(log[1/[H^+]^2]=6.085)#
From which:
#sf([H^+]=9.314xx10^(-4)color(white)(x)"mol/l")#
#sf(pH=-log[H^+]=-log[9.314xx10^(-4)])#
#sf(pH=3)#
