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)#