How to calculate the max pH at which reaction occurs using the Nernst equation?

For this reaction,

#"pH"_(max) = log{(["HOCl"]["Cl"^(-)])/(["Cl"_2(g)]) "exp"((nFE_(cell)^@)/(RT))}#

#~~ 0.4343ln((["HOCl"]["Cl"^(-)])/(["Cl"_2(g)])) + (0.4343nFE_(cell)^@)/(RT)#

where #n = 2#, and concentrations must be known ahead of time.

I derive this below.

The full reaction when adding these depends on what will make it spontaneous at #25^@ "C"#. I simply subtract the less positive #E^@# from the more positive #E^@#:

#E_(cell)^@ = E_(cathode)^@ - E_(anode)^@#

#= "1.63 V" - "1.359 V" = "0.27 V"#

where both values used are the given standard reduction potentials.

In this case, #"HOCl"(aq)# would spontaneously act as the oxidizing agent, and #"Cl"^(-)(aq)# gets oxidized to #"Cl"_2(g)#:

#2stackrel(color(blue)(+1))"H"stackrel(color(blue)(-2))"O"stackrel(color(blue)(+1))"Cl"(aq) + 2stackrel(color(blue)(+1))("H"^(+))(aq) + cancel(2e^(-)) -> stackrel(color(blue)(0))"Cl"_2(g) + 2stackrel(color(blue)(+1))"H"_2stackrel(color(blue)(-2))"O"(l)#
#ul(2stackrel(color(blue)(-1))("Cl"^(-))(aq) -> cancel(2e^(-)) + stackrel(color(blue)(0))("Cl"_2)(g)" "" "" "" "" ")#
#2stackrel(color(blue)(+1))"H"stackrel(color(blue)(-2))"O"stackrel(color(blue)(+1))"Cl"(aq) + 2stackrel(color(blue)(+1))("H"^(+))(aq) + 2stackrel(color(blue)(-1))("Cl"^(-))(aq) -> 2stackrel(color(blue)(0))"Cl"_2(g) + 2stackrel(color(blue)(+1))"H"_2stackrel(color(blue)(-2))"O"(l)#

And this we reduce down to:

#stackrel(color(blue)(+1))"H"stackrel(color(blue)(-2))"O"stackrel(color(blue)(+1))"Cl"(aq) + stackrel(color(blue)(+1))("H"^(+))(aq) + stackrel(color(blue)(-1))("Cl"^(-))(aq) -> stackrel(color(blue)(0))"Cl"_2(g) + stackrel(color(blue)(+1))"H"_2stackrel(color(blue)(-2))"O"(l)#

This reaction has the following reaction quotient:

#Q = (["Cl"_2(g)])/(["HOCl"]["H"^(+)]["Cl"^(-)])#

The Nernst equation relates #E_(cell)^@# to nonstandard states:

#E_(cell) = E_(cell)^@ - (RT)/(nF)lnQ#

where #RT# is known from, e.g. the ideal gas law (except #R = "8.314472 V"cdot"C/mol"cdot"K"# here), #n# is the number of electrons (#"mol e"^(-)"/mol atoms"#), #F = "96485 C/mol e"^(-)#, and #E_(cell)# is the cell potential.

The max pH for which the reaction "occurs" may mean the value for which the reaction ceases to be spontaneous, or reaches equilibrium, i.e. when #E_(cell) = 0#.

We should notice that

• since #"pH" = -log["H"^(+)]#, #"pH"# increases as #["H"^(+)]# decreases (less acidic).
• as #["H"^(+)]# decreases in the denominator of #Q#, we shall see that #Q# increases, which makes #-(RT)/(nF)lnQ# more negative (less positive).

Therefore, increasing #"pH"# for this reaction decreases #E_(cell)# from a more positive to a less positive value, and eventually would make the reaction nonspontaneous.

So, set #E_(cell) = 0# and rearrange:

#E_(cell)^@ = (RT)/(nF)lnQ#

#ln Q = (nFE_(cell)^@)/(RT)#

#Q = "exp"((nFE_(cell)^@)/(RT)) -= e^(nFE_(cell)^@//RT)#

#= (["Cl"_2(g)])/(["HOCl"]["H"^(+)]["Cl"^(-)])#

Knowing that #["H"^(+)] = 10^(-"pH")#:

#"exp"((nFE_(cell)^@)/(RT)) = (["Cl"_2(g)])/(["HOCl"]["Cl"^(-)]) cdot 10^("pH")#

Therefore, after rearranging, we get a general expression:

#"pH"_(max) = log{(["HOCl"]["Cl"^(-)])/(["Cl"_2(g)]) "exp"((nFE_(cell)^@)/(RT))}#

Or, if we want to simplify this further, note that #lnx ~~ 2.303logx#, so that:

#color(blue)("pH"_(max)) ~~ 1/(2.303) ln{(["HOCl"]["Cl"^(-)])/(["Cl"_2(g)]) "exp"((nFE_(cell)^@)/(RT))}#

#~~ color(blue)(0.4343ln((["HOCl"]["Cl"^(-)])/(["Cl"_2(g)])) + (0.4343nFE_(cell)^@)/(RT))#