Redox (II): Standard Electrode Potential Eo(5): ΔG, ΔS total, K and E⦵

Edexcel A level Chemistry (2017)

Topic 14: Redox (II): Standard Electrode Potential Eo(5)

Here is a further learning objective

14/11 To understand that Eocell is directly proportional to the total entropy change and to ln K for a reaction

Total entropy change, K and E^⦵

Let’s see in this post how the three thermodynamic functions ΔS_total, K and E^⦵  relate to each other.

We’ll begin our discussion looking at a typical reaction and we’ll calculate under standard conditions all the thermodynamic values we know that relate to this reaction.

The reaction we are going to look closely at is this one:

Zinc reduces copper ions to copper.  The reaction is spontaneous or feasible and we know that zinc is more reactive than copper.

Zn(s)   +    Cu²⁺ (aq)     ⟶    Zn²⁺(aq)     +     Cu(s)      at 298K

Let’s first calculate ΔH^o₂₉₈ using this equation:

ΔH^o_{298   =}     ΔH^o_f (products)    —    ΔH^o_f (reactants)

ΔH^o_{298   =}     –153.9    —    +64.8

ΔH^o_{298   =}     –218.7kJ.mol^—1

Now we should be able to calculate the entropy changes in the system and the surroundings.

            ΔS^o_{sys   =}     S^o₂₉₈ (products)    —    S^o₂₉₈(reactants)

ΔS^o_{sys   =}     (–112.1 + +33.2)    —  ( +41.6 + –99.6)

ΔS^o_{sys   =}    —79.2    —    —58

ΔS^o_{sys   =}     —21.2 J.mol^—1K^—1

And in the surroundings:

ΔS^o_surroundings   =  —ΔH^o / T

ΔS^o_surroundings   =  218,700 / 298

ΔS^o_surroundings   =  +734 J.mol^—1K^—1

So I hope you can see that the negative entropy change in the system is offset since the entropy change in the surroundings is so much more positive — by about +500 J.mol^—1K^—1

Therefore we can easily work out

ΔS^o_{total   =}     ΔS^o_sys     +    ΔS^o_surroundings

ΔS^o_{total   =}     –21.2     +    +734

ΔS^o_{total   =}     +712.8 J.mol^—1K^—1

Now you should see the implications of this highly positive value for the total entropy change.  The value implies that the chemical change is feasible at 298K.

But there is something else here that is even more important.

You’ll see that in this redox reaction there was a total entropy increase of 734 J.mol^—1K^—1 and 218.7kJ of heat were transferred to the surroundings.

But looking closely at the data it wasn’t necessary to transfer all that heat to the surroundings to compensate for the negative entropy change in the system.

We only needed to transfer enough heat to stop the total entropy decreasing i.e. enough heat to provide an entropy change in the surroundings of about +21.2 J.mol^—1K^—1  

How much heat would be required we could work out because we know that

ΔS^o_surroundings   =  —ΔH^o / T

or

ΔH^o  =  TΔS^o_surroundings

or

ΔH^o  =  298 × 21.2  =   6317J or 6.317 kJ.

This small value of energy would compensate for the decrease in entropy of the system.

The rest of the energy released could do useful work and we could calculate what the value of that work would be:  

218—6.3 kJ = 211.7kJ

This is the maximum amount of work we can obtain from this chemical reaction.

Let’s generalise our discussion.

At constant temperature and pressure we can say that the maximum amount of useful work we can obtain from a chemical change is given by

—(ΔH — TΔS)

or

w_max   =   —ΔG

So the maximum amount of useful work we obtain from a chemical change is another way of talking about the Gibbs free energy change (ΔG) for that reaction.

Now if we set up the Daniell Cell in which the reaction of zinc and copper ions generates an emf of +1.1v then we find that the maximum amount of useful work is related to the cell emf:

w_max   =   —ΔG  = nFE

where n is the number of moles of electrons involved:  2

F is the Faraday Constant:  96000Coulombs/mol.

E is the standard cell potential: 1.10v

If add these values to the equation we obtain a value for —ΔG

—ΔG = 211.2kJ.mol^—1

This value (211.2kJ) for —ΔG from the cell potential value is almost the same as the value (211.7kJ) we obtained from measuring the enthalpy and entropy changes for the zinc/copper ion reaction.

Finally in a previous post we determined the relationship between the free energy change ΔG and the equilibrium constant Kc.

—ΔG  =  RT ln K_c

So what you can do is use the value for —ΔG to determine K.

—ΔG  =  RT ln K_c

so

211.2 × 1000  =   8.314 × 298 ln K

you multiply by 1000 to rationalise the units from kiloJoules into Joules,

you use 298K as the standard conditions temperature,

you use 8.314 J.mol^—1.K^—1 which is R the gas constant.

Then you can calculate K as 1.05  ×  10³⁷

K is a huge number!

Let’s summarise what we have learnt so for.

For a reaction that is spontaneous and feasible we have calculate several thermodynamic quantities

Equilibrium constant K_c =  1.05  ×  10³⁷

Standard cell potential E_cell = 1.1v

Standard entropy change  ΔS  =  +793 J.mol^—1.K^—1  

Standard free energy change  ΔG  = —211.7kJ.mol^—1

These values are for a specific reaction and give you some idea of how you can use them to determine whether another reaction is also spontaneous and feasible or not.

You can work through a similar set of calculations to find the criteria for non-spontaneous reactions etc.

If you do those calculations then this table shows you the kind of values you will come up with:

Value of	Reaction does not ‘go’, it is not feasible	Reactants dominate in the equilibrium	Equal amounts of products and reactants: a “perfect” equilibrium	Products dominate in an equilibrium	Reaction goes to completion i.e. it is feasible or spontaneous
Kc	<10—10	10—2	Kc = 1	102	>10—10
ΔGo	> +60	+10	ΔGo = 0	—10	< —60
ΔSo	<—200	—40	ΔSo = 0	+40	> +200
Eocell	<—0.6	—0.1	Eocell = 0	+0.1	> 0.6

You can use these values to predict whether a reaction will take place or not because each is proportional to the Gibbs free energy change:

ΔG^o   ∝ E^o_cell  since  —ΔG^o   = nFE^o_cell     

ΔG^o   ∝ ΔS^o    since  ΔG^o   = ΔH^o   — TΔS^o      

and

ΔG^o   ∝ K_c    since  —ΔG^o   = RT ln K_c