Free Energy & Eº (DP IB Chemistry: HL)

• Previously we have seen the concept and term free energy, ΔG^θ
• Free energy is a measure of the available energy to do useful work and takes into account the entropy change, ΔS^θ, as well as the enthalpy change of a reaction, ΔH^θ
• For reactions to be spontaneous, the free energy change must be negative
• We have also seen that to calculate a cell potential using standard electrode potentials we use the expression:

EMF = E_RHS – E_LHS

• This is not an arbitrary arrangement of the terms
• The convention of placing the half cell with the greatest negative potential on the left of the cell diagram ensures that you will always get a positive reading on the voltmeter, corresponding to the spontaneous reaction
  • If you have done an experiment on measuring electrode potentials, you have probably been told to 'swap the terminals if you don't get a positive reading on the voltmeter'

• In electrochemical cells, a spontaneous reaction occurs when the combination of half cells produces a positive voltage through the voltmeter, i.e. the more negative electrode pushes electrons onto the more positive electrode
• It doesn't really matter if you are using a digital multimeter as a voltmeter as you will still get a reading (with the wrong sign), but analogue voltmeters will only work if the terminals are correctly connected to the positive and negative half cells
• This should give you an insight into why the following statements are true:

If ΔE^θ is positive, the reaction is spontaneous as written

If ΔE^θ is negative, the forward reaction is non-spontaneous but the reverse reaction will be spontaneous

• You should now be able to see that there is a link between  ΔG^θ and E^θ
• This relationship is the equation:

ΔG^θ = -nFE^θ

where:

• • n= number of electrons transferred
  • F= the Faraday constant, 96 500 C mol^-1

• When a reaction has reached equilibrium, there is no free energy change so ΔG^θ  is zero and it follows that E^θ must also be zero
• This is effectively what happens when the reactants in a voltaic cell have been exhausted and there is no longer any push of electrons from one half cell to the other

Answer

• • Write the equation:

ΔG^θ =-nFE^θ

• • Substitute the values

ΔG^θ = - 2 x 96 500 C mol^-1 x 1.10 V =- 212300 C mol^-1 V

• • This looks a strange unit. However, by definition 1J = 1V x 1C, so this answer can be expressed as

ΔG^θ = - 212300 J mol^-1 or  -212.3 kJ mol^-1

• The three conditions of free energy and electrode potential are summarised below

Summary table of the conditions of free energy and electrode potential

• To predict the spontaneous reaction, we simply need to find the relevant half equations and electrode potentials
• From this information, we can deduce the spontaneous and non-spontaneous reaction
• By using the convention:

EMF = E_RHS – E_LHS

• A positive EMF is obtained from the spontaneous reaction which occurs when the most negative half cell is E_LHS and the most positive is E_RHS
• The left side is always where oxidation takes place so we can also us an alternative form of the relationship:

EMF = E_reduction – E_oxidation

Answer

• • Table 24 of the Data Book shows the following half reactions

Sn²⁺ (aq) + 2e^-  → Sn (s)          E^θ = -0.14 V

Mn²⁺ (aq) + 2e^-  → Mn (s)         E^θ = -1.18 V

• • Manganese is the more negative value, so will be E_LHS or E_oxidation in the spontaneous reaction  

EMF =E_RHS – E_LHS = (-0.14) - (-1.18) = +1.04

• • For oxidation to take place, the manganese must lose electrons and the tin(II) must gain electrons

Mn (s) → Mn²⁺ (aq) + 2e^-    and   Sn²⁺ (aq) + 2e^- → Sn (s) 

• • So, the spontaneous reaction is

Mn (s) +  Sn²⁺ (aq) → Mn²⁺ (aq) +  Sn (s)

• • Therefore, the reaction in the question is not spontaneous