A mercury electrode is used to measure the #["Hg"^(2+)]# in a test solution. The electrode is measured to have a potential of #"0.793 V"#. Calculate the concentration of #["Hg"^(2+)]#?

I got #"0.0118 M"#.

The idea is that the measured potential is at a nonstandard condition, and so, we consider

#DeltaG = DeltaG^@ + RTlnQ#.

There exists the relationship

#DeltaG = -nFE#,

so

#-nFE = -nFE^@ + RTlnQ#,

giving us the Nernst equation:

#E = E^@ - (RT)/(nF)lnQ#,

where:

• #E# is the voltage difference measured at a nonstandard temperature and pressure. #@# indicates #"1 atm"# and #25^@ "C"#.
• #R = "8.314472 V"cdot"C/mol"cdot"K"# is the universal gas constant.
• #T# is the temperature in #"K"#.
• #n# is in units of #"mol e"^(-)/("mol reactant")#.
• #F = "96485 C/mol e"^(-)# is the Faraday constant.
• #Q# is the reaction quotient.

So, we can first calculate the reaction quotient, i.e. the non-equilibrium constant.

#=> E^@ - E = (RT)/(nF)lnQ#

#=> (nF)/(RT)(E^@ - E) = lnQ#

The reaction quotient is then:

#Q = "exp"[(nF)/(RT)(E^@ - E)]#

Since you didn't state any temperature, I assume standard temperature but nonstandard concentrations and/or pressure...

#Q = "exp"[(2 cancel("mol e"^(-))//cancel("mol Hg"^(2+)) cdot 96485 cancel"C""/"cancel("mol e"^(-)))/(8.314472 cancel"V"cdotcancel"C""/"cancel"mol"cdotcancel"K" cdot 298.15 cancel"K") cdot (0.850 cancel"V" - 0.793 cancel"V")]#

#= 84.53#

The reaction quotient for this reaction is then, noting that mercury is a LIQUID and not a solid:

#Q = (color(red)(cancel(color(black)(["Hg"(l)]))^(color(black)(1))))/(["Hg"^(2+)]) = 84.53#

#=> color(blue)(["Hg"^(2+)] = "0.0118 M")#

The potential of a half - cell is given by:

#sf(E=E^(@)-(RT)/(zF)ln([["reduced form"]]/[["oxidised form"]))#

For the #sf(Hg_((aq))^(2+)"/"Hg_((l))# half - cell we take #sf([Hg_((l))])# to be constant so that at #sf(25^(@)C)# the expression simplifies to:

#sf(E=E^(@)+0.0591/(2)log[Hg^(2+)])#

#:.##sf(0.793=0.8501+0.0591/(2)log[H^+])#

#sf((-0.0572xx2)/(0.0591)=log[Hg^(2+)])#

#sf(log[Hg^(2+)]=-1.9289)#

#sf([Hg^(2+)]=0.0118color(white)(x)"mol/l")#