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)
