Determine of `pH`: The following cell has a potential of `0.55 V` at `25^(@)C`:
`Pt(s)|H_(2)(1 bar)|H^(+)(aq. ? M)||Cl^(-)(1 M)|Hg_(2)Cl_(2)(s)|Hg(l)`
What is the `pH` of the solution in the anode compartment?
Strategy: First, read the shorthand notation to obtain the cell reaction. Then, calculate the half cell potential for the hydrogen electrode from the observed cell potential and the half cell poten-tial for the calomel reference electrode. Finally, apply the Nernst equation to find the `pH`.

Explanation

The cell reaction is
`H_(2)(g)+Hg_(2)Cl_(2)(s)hArr 2H^(+)(aq.)+2Hg(l)+2Cl^(-)(aq.)`
and the potential is
`E_("cell") = E_("anode") + E_("cathode")`
`E_("cell") = E_(H_(2)//H^(+)) + E_(Hg_(2)Cl_(2)//Hg = 0.55 V`
Because the reference electrode is the standard calomel electode, whic has `E = E^(@) = 0.28 V`, the half-cell potential for the hydrogn electode is `0.27 V`:
`E_(H_(2)//H^(+)) = E_("cell") - E_(Hg_(2)Cl_(2)//Hg`
`= (0.55 V) - (0.28 V)`
`= 0.27 V`
Applying the Nernst equation to the half-reaction
`H_(2)(g) rarr 2H^(+)(aq.) + 2e^(-)`:
`E_(H_(2)//H^(+)) = E_(H_(2)//H^(+))^(@) - ((0.0592 V)/(n)) "log" ((C_(H^(+))^(2))/(P_(H_(2))))`
Substituting in the values of `E, E^(@)`, n and `P_(H_(2))` gives
`0.27 V = (0V) - ((0.0592 V)/(2)) "log"((C_(H+)^(2))/(1))`
`= (0.0592 V) (pH)`
Therefore, the `pH` is
`pH = (0.27 V)/(0.0592 V) = 4.6`