What are the total enthalpy changes for each of the following oxidation or reduction processes?

#"Cl"^(-)(g) -> "Cl"^(+)(g) + 2e^(-)#
#2e^(-) + "K"^(+)(g) -> "K"^(-)(g)#

1 Answer
Truong-Son N.
May 30, 2017

#DeltaH_("tot","Cl"^(-)->"Cl"^(+)) = DeltaH_(IE,"Cl"(g)) - DeltaH_(EA,"Cl"(g)) = "1599.8 kJ/mol"#
#DeltaH_("tot","K"^(+)->"K"^(-)) = DeltaH_(EA,"K"(g)) - DeltaH_(IE,"K"(g)) = -"467.2 kJ/mol"#

Since you are in the gas phase, we are talking about ionization energies and electron affinities. Recall:

IONIZATION ENERGY

It is the energy input in order to remove an electron.

Ionization is therefore described by:

#A(g) -> A^(+)(g) + e^(-)#, #" "DeltaH_(IE) = . . . #

or similar.

Since enthalpy is a state function, #-DeltaH# is for the reverse of the process described by #DeltaH#.

ELECTRON AFFINITY

It is the energy change due to adding an electron. If it's negative, we can imagine that the atom gets more stable. If it's positive, the atom got less stable (hence, the noble gases have nonnegative electron affinities).

Electron affinity is therefore described by:

#A(g) + e^(-) -> A^(-)(g)#, #" "DeltaH_(EA) = . . . #

or similar.

Now, for the following processes...

CHLORINE PROCESS

#a)# #"Cl"^(-)(g) -> "Cl"^(+)(g) + 2e^(-)#

We break this into two steps. (We know the values for starting at the neutral atom, but not the ions.)

#"Cl"^(-)(g) -> "Cl"(g) + e^(-)# #" "bb((1))#, #" "-DeltaH_(EA,"Cl"(g)) = . . . #

#"Cl"(g) -> "Cl"^(+)(g) + e^(-)# #" "bb((2))#, #" "DeltaH_(IE,"Cl"(g)) = . . . #

But as noted above, we recognize #(1)# is the reverse of the electron affinity for #"Cl"(g)#.

Hence, by Hess's law, we add #(1)# and #(2)# according to #DeltaH_"tot" = DeltaH_1 + . . . + DeltaH_N#, or:

#DeltaH_"tot" = DeltaH_(IE) - DeltaH_(EA)#

We look these up to be:

#DeltaH_(IE,"Cl"(g)) = "1251.2 kJ/mol"#, ref

#DeltaH_(EA,"Cl"(g)) = -"348.6 kJ/mol"#, ref

So,

#color(blue)(DeltaH_("tot","Cl"^(-)->"Cl"^(+))) = 1251.2 - (-348.6) = color(blue)("1599.8 kJ/mol")#

POTASSIUM PROCESS

#b)# #"K"^(+)(g) + 2e^(-) -> "K"^(-)(g)#

Similar process as before, breaking into two steps. (We know the values for starting at the neutral atom, but not the ions.)

#"K"^(+)(g) + e^(-) -> "K"(g)# #" "bb((1))#, #" "-DeltaH_(IE,"K"(g)) = . . . #

#"K"(g) + e^(-) -> "K"^(-)(g)# #" "bb((2))#, #" "DeltaH_(EA,"K"(g)) = . . . #

We look these up to be:

#DeltaH_(IE,"K"(g)) = "418.8 kJ/mol"#, ref

#DeltaH_(EA,"K"(g)) = -"48.4 kJ/mol"#, ref

So, by Hess's law again:

#color(blue)(DeltaH_("tot","K"^(+)->"K"^(-))) = DeltaH_(EA,"K"(g)) - DeltaH_(IE,"K"(g))#

#= -48.4 - (418.8) = color(blue)(-"467.2 kJ/mol")#

If you notice, these processes were opposite in direction.

#DeltaH_(IE) - DeltaH_(EA) = -(DeltaH_(EA) - DeltaH_(IE))#

The left-hand side was the chlorine process, and the right-hand side (without the negative sign) was the potassium process. So, it makes some sense that they are opposite signs.

Related questions
See all questions in Enthalpy
Impact of this question
1793 views around the world
You can reuse this answer
Creative Commons License