IB Colourful Solutions in Chemistry

Both electrode potentials and Gibbs free energy can be used to predict the spontaneity of a reaction.

Reactivity 3.2.14 - The equation ΔG^⦵ = − nFE^⦵cell shows the relationship between standard change in Gibbs energy and standard cell potential for a reaction. (HL)

Determine the value for ΔG^⦵ from E^⦵ data.

Guidance

The equation and the value of F in C mol^–1 are given in the data booklet.

Tools and links

Reactivity 1.4 - How can thermodynamic data also be used to predict the spontaneity of a reaction?

Gibb's free energy
Worked examples

Gibb's free energy

For any reaction to proceed the Gibb's Free Energy change, ΔG^o (a measure of the entropy of the universe) must have a negative value. It follows that as the requirement for spontaneity of a reaction according to electrode potentials is for the value of E^o for a redox reaction to be positive then this must be related to a negative Gibbs Free Energy change.

ΔG^o = -nE^o

where 'n' is a proportionality constant.

It turns out that ΔG^o depends on the total electrical charge transferred during reaction. This may be given by the number of moles of electrons, z, and the total charge on one mole of electrons transferred during the redox process. Thus:

ΔG^o = -zFE^o

Where 'z' is the moles of electrons transferred and 'F' is the charge on one mole of electrons (1 Faraday = 96,500 Coulombs approx).

This equation allows calculation of the Gibb's Free Energy change from a consideration of the electrode potentails of the redox components.

Example: Find the Gibb's Free Energy of the reaction between a solution containing copper 2+ ions (1.0 mol dm^-3) and a sample of zinc metal at 25^oC.

Cu²⁺ + 2e ⇋ Cu		E^o = +0.34V
and
Zn²⁺ + 2e ⇋ Zn		E^o = -0.76V
then
E = E (red) - E (ox)		E = +0.34 - (-0.76) V
		E = + 1.10V

ΔG^o = -zFE^o
and
z = 2 moles of electrons, F = 96,500 Coulombs
Gibbs Free Energy =		- 212.300 kJ

It should be remembered that this value only applies to standard conditions and that any change in conditions will cause a consequent change in the value of both E and ΔG

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Worked examples

R3214-1 For a thermodynamically spontaneous cell reaction, which statement is correct?

E^o (reduction) must be more negative than E^o (oxidation) by 0.3V.
E^o (oxidation) must be more negative than E^o (reduction) by 0.3V.
E^o(cell) should be negative.
The difference between E^o (reduction) and E^o (oxidation) must be more than 1.0 V.

Answer

Normally we use the value of +0.3V to indicate whether a reaction is spontaneous or not. The oxidation process refers to the removal of electrons, i.e. the state getting oxidised. he equation used is:

E^o = E^o(red) - E^o(ox)

For E^o to be positive by more than 0.3V, then E^o(ox) must be more negative than E^o(red) by 0.3V.

R3214-2 Which signs are correct for a spontaneous reaction occurring in a cell?

E^o positive and delta G^o positive
E^o positive and delta G^o negative
E^o negative and delta G^o positive
E^o negative and delta G^o negative

Answer

Spontaneous reaction has E^o positive and ΔG^o negative.

R3214-3 An unknown ion, MO₃^- can be converted in acidic solution into an ion, M²⁺, according to the redox half-equation:

MO₃^- + 6H⁺ + 3e → M²⁺ + 3H₂O E^o = 1.20 V

Using the electrode potential data explain how this value could be used to predict whether MO₃^- would oxidise Br^- ions to Br₂ or whether Br₂ would oxidise colourless M²⁺ to MO₃^-. Write a balanced equation for the reaction that would be expected to occur.

Answer

From the table, E^o for Br₂(l)|Br^-(aq) is 1.07 V. This is less positive than the E^o value for MO₃^-(aq)|M²⁺(aq). This means that MO₃^-(aq) is a better oxidising agent than Br₂(l).

Hence MO₃^-(aq) oxidises Br^-(aq) to bromine.

Before the equation can be constructed the number of electrons must first be equalised:

1 - MO₃^- + 6H⁺ + 3e → M²⁺ + 3H₂O

2 - 2Br^-(aq) → Br₂(l) + 2e

Multiply through by 2 in equation 1 and multiply through by 3 in equation 2

3 - 2MO₃^- + 12H⁺ + 6e → 2M²⁺ + 6H₂O

4 - 6Br^-(aq) → 3Br₂(l) + 6e

Add 3 + 4 gives: 2MO₃^- + 12H⁺ + 6Br^-(aq) → 2M²⁺ + 6H₂O + 3Br₂(l)

R3214-4 Using the value for the standard electrode potential of copper, +0.34V, determine whether, or not, there is a spontaneous reaction between copper metal and a solution containing hydrogen ions.
Answer

Hydrogen ions (at standard concentration) in aqueous solution have an electrode potential of 0 V.

If copper were to react, it would become copper(II), i.e. it would be oxidised. Using the equation:

E = E(reduced species) - E(oxidised species) = 0 - 0.34 = -0.34 V

This is a negative value, so the reaction is not spontaneous, i.e. it will not happen (under standard conditions).

R3214-5 Using the following standard electrode potential data, identify a substance that will oxidise bromide ions, but not chloride ions. Explain your choice and write an equation for the redox reaction that you have chosen.
Answer

For a substace to oxidise bromide ions it must have an electrode potential more positive than that of bromine. To not be able to oxidise chloride ions, it must have a potential less positive than chlorine. From the list given, the only species with an electrode potential between that of chlorine and bromine is oxygen in acidic solution.

R3214-6 A cell was set up using zinc in zinc sulfate solution and copper in copper(II) sulfate solution, both solutions under standard conditions. Using the following standard electrode potential data calculate the cell potential and write an equation for the spontaneous cell reaction.
Answer

The standard cell potential is given by the difference in standard electrode potentials of the half cells = 0.34 -- 0.76 = 1.10 V.

Zinc is the better reducing agent, thus the half-cell equations for the spontaneous process are:

Zn(s) → Zn²⁺(aq) + 2e^-
Cu²⁺(aq) + 2e^- Cu(s)

These can be added directly as the number of electrons involved in each is the same.

Cu²⁺(aq) + Zn(s) Zn²⁺(aq) + Cu(s)

R3214-7 Chlorine gas is formed when potassium manganate(VII) is added to concentrated hydrochloric acid. Calculate the cell potential for this reaction and deduce the equation for the reaction. Using the following standard electrode potential data, explain why potassium dichromate(VI) does not react with concentrated hydrochloric acid.
Answer

The cell potential for the reaction between hydrochloric acid and potassium manganate(VII) is given by the difference in electrode potentials = 1.49 - 1.36 V = 0.13 V

Chloride ions are oxidised according to the equation 1:

1 2Cl^- → Cl₂ + 2e

Manganate ions react according to the equation 2:

2 MnO₄^-(aq) + 8H⁺ + 5e^- → Mn²⁺(aq) + 4H₂O

The equations can only be added after the electrons are equalised. Multiply through by 5 in equation 1 and multiply through by 2 in equation 2

3: 10Cl^- → 5Cl₂ + 10e

4: 2MnO₄^-(aq) + 16H⁺ + 10e^- 2Mn²⁺(aq) + 8H₂O

Add equations 3 and 4

2MnO₄^-(aq) + 16H⁺ + 10Cl^- → 2Mn²⁺(aq) + 8H₂O + 5Cl₂

Potassium dichromate does not react with hydrochloric acid as the electrode potential is less positive than that of chlorine, therefore the dichromate ion cannot oxidise chloride ions (under standard conditions).

R3214-8 From the following data:

Zn²⁺(aq) | Zn E^o = -0.76 V
Fe³⁺(aq), Fe²⁺(aq) | Pt E^o = +0.77 V

it can be deduced that:

I. The standard E.M.F. for the cell Zn|Zn²⁺(aq) ¦ Fe³⁺(aq), Fe²⁺(aq)|Pt is 0.01 V.
II. Zinc is a more powerful reductant than Fe²⁺ ions.
III. Fe³⁺ can oxidize zinc under standard conditions.

I, II and III are correct.
I and II are correct.
II and III are correct.
I is the only correct response.
III is the only correct response.

Answer

Statement I is incorrect, the cell potential is given by the difference between the half-cell potentials = 1.53 V.

The most powerful reducing agent has the most negative potential, in this case Zn(s).

Statement III is correct as Fe³⁺ is the oxidising agent and Zn(s) is the reducing agent.

R3214-9 Given the standard electrode (reduction) potentials:

Ni²⁺(aq) + 2e → Ni(s) E° = - 0.23 V

Cr³⁺(aq) + 3e → Cr(s) E° = - 0.74 V

Which pair of substances will react spontaneously?

Ni²⁺ with Cr³⁺
Ni with Cr³⁺
Ni²⁺ with Cr
Ni with Cr

Answer

The most powerful oxidising agent is Ni²⁺ (most positive on the left hand side). The best reducing agent is Cr(s) (most negative on the right hand side).

Therefore the spontaneous reaction is between Ni²⁺ and Cr

R3214-10 From the given standard electrode potentials which statement is correct:

Ca²⁺(aq) + 2e → Ca(s) E^o = -2.87 V

Ni²⁺(aq) + 2e → Ni(s) E^o = -0.23 V

Fe³⁺(aq) + e → Fe²⁺(aq) E^o = +0.77 V

Ca²⁺(aq) can oxidise Ni(s)
Ni²⁺(aq) can reduce Ca²⁺(aq)
Fe³⁺(aq) can oxidise Ni(s)
Fe³⁺(aq) can reduce Ca²⁺(aq)

Answer

The most negative potential shows the best reducing agent, in this case Ca(s). The best oxidising agents are the ions with the most positive potential, in this case Fe²⁺(aq).

The first choice is not possible as Ni is less reactive than Ca²⁺(aq). The second choice is not possible, as both of the reactants appear on the same side of the redox list. The third choice has Ni(s), which is a more negative than the Fe³⁺(aq), the oxidising side of the redox equilibria. Hence, Fe³⁺(aq) can oxidise Ni(s).

R3214-11 Consider the following reactions:

Cu²⁺(aq) + 2e → Cu(s) E^o = +0.34 V

Mg²⁺(aq) + 2e → Mg(s) E^o = -2.36 V

Zn²⁺(aq) + 2e → Zn(s) E^o = -0.76 V

Which statement is correct?

Cu²⁺(aq) will oxidise both Mg(s) and Zn(s)
Zn(s) will reduce both Cu2+(aq) and Mg²⁺(aq)
Mg²⁺(aq) will oxidise both Cu(s) and Zn(s)
Cu(s) will reduce both Zn²⁺(aq) and Mg²⁺(aq)

Answer

Magnesium is the best reducing agent (most negative E^o value) and copper ions are the best oxidising agent. Out of the choices, the only true statement is that Cu²⁺(aq) will oxidise both Mg(s) and Zn(s)

R3214-12 Consider the standard electrode potentials of the following reactions

Cr³⁺(aq) + 3e → Cr(s) E^o= -0.75 V
Cd²⁺(aq) + 2e → Cd(s) E^o= -0.40 V

What is the value of the cell potential in volts for the following reaction?

2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)

0.35
1.15
-0.30
-0.35

Answer

Standard cell potentials are obtained by the difference between the standard electrode potentials.

In this case -0.40 - (-0.75) = +0.35 V

R3214-13 The standard electrode potentials for two half-cells involving iron are given below.

Fe²⁺(aq) + 2e → Fe(s) E^o = -0.44V
Fe³⁺(aq) + 1e → Fe²⁺(aq) E^o = +0.77

What is the equation and the cell potential for the spontaneous reaction that occurs when the two half cells are connected?

3Fe²⁺(aq) → Fe(s) + 2Fe³⁺(aq) E^o = +1.21 V
Fe²⁺(aq) + Fe³⁺(aq) → 2Fe(s) E^o = +0.33 V
Fe(s) + 2Fe³⁺(aq) → 3Fe²⁺(aq) E^o = +0.33 V
Fe(s) + Fe³⁺(aq) → 3Fe²⁺(aq) E^o = +1.21 V

Answer

A spontaneous reaction occurs between the best oxidising agent (most positive on the left hand side) and the best reducing agent (most negative on the right hand side). In this case it is the reaction between Fe³⁺(aq) and Fe(s). To obtain the balanced equation the electrons must be equalised before the equations are added together.

Equation 1: Fe(s) → Fe²⁺(aq) + 2e

Equation 2: Fe³⁺(aq) + 1e → Fe²⁺(aq)

Multiply through by 2 in equation 2 → equation 3: 2Fe³⁺(aq) + 2e → 2Fe²⁺(aq)

Add equations 1 and 3 : Fe(s) + 2Fe³⁺(aq) → Fe²⁺(aq) + 2Fe²⁺(aq)

Collect terms to give the final equation: Fe(s) + 2Fe³⁺(aq) → 3Fe²⁺(aq)

R3214-14 The standard electrode potentials of three elements are as follows.

X: +1.09 V
Y: +0.54 V
Z: +1.36 V

Which statement is correct?

Z will oxidise Y^-(aq) and X^-(aq)
Y will oxidise Z^-(aq) and X^-(aq)
X will oxidise Y^-(aq) and Z^-(aq)
Z will oxidise Y^-(aq) but not X^-(aq)

Answer

Of the three elements Z (the most positive) is the best oxidising agent and will oxidise Y^-(aq) and X^-(aq)

R3214-15 Using the standard electrode potentials below, determine which of the following statements is correct.

Sn⁴⁺(aq) + 2e → Sn²⁺(aq) E^o = +0.15V
Fe³⁺(aq) + 1e → Fe²⁺(aq) E^o = +0.77V

Fe²⁺(aq) can spontaneously reduce Sn⁴⁺(aq)
Sn²⁺(aq) is a better reducing agent than Fe²⁺(aq)
These two half cells can be put together to produce a cell with E^o = 0.92V
The Fe³⁺(aq) / Fe²⁺(aq) E^o value must be multiplied by two when calculating the cell voltage for a reaction between Fe³⁺(aq) and Sn²⁺(aq)

Answer

The spontaneity of a reaction can be predicted using the formula:

E = E(reduced species) - E(oxidised species)

From consideration of the electrode potentials Fe³⁺(aq) will oxidise Sn²⁺(aq). This, however is not one of the choices.

The most negative (least positive) species is Sn²⁺(aq) meaning that it is the best reducing agent from the choices.

Therefore Sn²⁺(aq) is a better reducing agent than Fe²⁺(aq)

R3214-16 Consider the standard electrode potentials of the following reactions:

Sn⁴⁺(aq) + 2e → Sn²⁺(aq) E^o = 0.15V
Fe³⁺(aq) + 1e → Fe²⁺(aq) E^o = 0.77V

What is the value of the cell potential (in volts) for the spontaneous reaction?

+1.69
+1.39
+0.92
+0.62

Answer

The cell potential is obtained by the difference between the two standard electrode potentials of the half-cells.

Therefore standard cell potential = 0.77V - 0.15V = 0.62V

R3214-17 Two half-equations and their standard electrode potentials are shown in the table below:

half-equation	E^o/V
Pb²⁺(aq) + 2e → Pb(s)	- 0.13
Ag⁺(aq) + 1e → Ag(s)	+ 0.80

What is the cell potential in Volts for the following reaction? Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s)

0.67
0.93
1.47
1.73

Answer

The cell potential is obtained by the difference between the two standard electrode potentials of the half-cells.

Therefore standard cell potential = 0.80V - (-0.13V) = 0.93V

R3214-18 Consider the standard redox potentials for the reactions shown:

Sn²⁺ + 2e^- → Sn +0.15 V
I₂ + 2e^- → 2I^- +0.54 V
Ag⁺ + e^- → Ag +0.8O V
Cl₂ + 2e^- → 2Cl^- +1.36 V
Au³⁺ + 3e^- → Au +1.50 V

Which one of the following is true?

silver will displace gold from a solution of gold ions
tin ions will displace chlorine from a solution of chloride ions
iodide ions will displace tin from a solution of tin ions
iodine will displace chlorine from a solution of chloride ions

Answer

Comparing the standard electrode potentials of silver and gold we see that silver is more negative (less positive) than gold and so silver metal will displace gold from a solution of gold ions.

The standard electrode potential tin is less positive than that of chlorine, so that the oxidising nature of the left hand side, i.e. Sn²⁺ is weaker than Cl_2. Thus tin ions will not displace chlorine.

The standard electrode potential of iodide ions is more positive that that of tin so iodide ions are a poorer reducing agent than tin and cannot reduce tin ions to tin.

The standard electrode potential of iodine is less positive that that of chlorine, so iodine cannot displace (oxidise) chloride ions to chlorine.

R3214-19 The cell represented below was set up under standard conditions.

Pt|H₂SO₃(aq),SO₄^2-(aq)||MnO₄^-(aq),Mn²⁺(aq)|Pt

Using the following standard electrode potential data, calculate the e.m.f. of this cell and write an equation for the spontaneous cell reaction.

Answer

Conventionally, the emf of a cell is calculated by E^o(right hand side) - E^o (left hand side)

Thus, the emf of the cell = 1.49 - = V

Inspection of the electrode potentials shows us that the manganate (VII) ion is the best oxidising species and H2SO3 is the best reducing species. Hence reaction is between this pair.

equation 1: H₂SO₃(aq) + H₂O → SO₄^2-(aq) + 4H⁺ + 2e

equation 2: MnO₄^-(aq) + 8H⁺ + 5e^- → Mn²⁺(aq) + 4H₂O

Before the equations can be added the electrons must be equalised. This is done by multiplying through by 5 in equation 1, and mutiplying through by 2 in equation 2:

equation 3: 5H₂SO₃(aq) + 5H₂O → 5SO₄^2-(aq) + 20H⁺ + 10e

equation 4: 2MnO₄^-(aq) + 16H⁺ + 10e^- → 2Mn²⁺(aq) + 8H₂O

Now add 3 and 4:

5H₂SO₃(aq) + 5H₂O + 2MnO₄^-(aq) + 16H⁺ → 5SO₄^2-(aq) + 20H⁺ + 2Mn²⁺(aq) + 8H₂O

Now cancel out similar terms on both sides:

5H₂SO₃(aq) + 2MnO₄^-(aq) → 5SO₄^2-(aq) + 4H⁺ + 2Mn²⁺(aq) + 3H₂O

R3214-20 Blocks of magnesium are bolted onto the hulls of iron ships in an attempt to prevent the iron being converted into iron(II), one of the steps in the rusting process. Use the standard electrode potential data, where appropriate, to calculate the e.m.f. of the cell represented by Mg(s)|Mg²⁺(aq)||Fe²⁺(aq)|Fe(s) under standard conditions, and write an equation for the reaction proceeding at the negative electrode of the cell when current flows.
Answer

Magnesium is a better reducing agent than iron (more negative electrode potential), so magnesium reacts with iron 2+ ions according to the equation:

Mg(s) + Fe²⁺(aq) → Fe(s) + Mg²⁺(aq)

The potential of this cell is (RHS - LHS) = -0.44 -- 2.37 = 1.93 V

The negative electrode produces electrons as the magnesium dissolves:

Mg(s) → Mg²⁺(aq) + 2e

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