14 Redox Equilibria

14.1 Redox Equations (Review)14.2 Electrode Potentials and the Electrochemical Series14.3 Predicting the Direction of Redox Reactions14.4 Uses for Electrochemical Cells

14.1 Redox Equations

Learning Objectives:1. Calculate the oxidation states of different elements in a chemical

reaction.2. Identify which species have been oxidised and which have been

reduced in a redox reaction.3. Write balanced redox equations.

Rules for Finding Oxidation Numbers

1. Uncombined elements = 02. Hydrogen = +1 (except metal hydrides = -1)3. Group 1 = +14. Group 2 = +25. Aluminium = +36. Oxygen = -2 (except peroxides = -1, compounds with F = +2)7. Fluorine = -18. Chlorine = -1 (except compounds with F and O = +)

Oxidation States Balance Out to Equal Charge of the Molecule

Balancing Redox Equations

1. Write balanced half equations. Remember H+ or H2O molecules may need to be added to balance.

2. Balance the electrons.3. Add the half equations and cancel anything appearing on both

sides.

Ex: Chlorine will oxidise iron(II) to iron(III) and is itself reduced to chloride ions.

Step 1: Write half equations

Cl2 (g) + 2e- 2Cl-(aq) reduction (0 -1)

Fe2+ (aq) Fe3+

(aq) + e- oxidation (+2 +3)

Step 2: Balance for electrons

Cl2 (g) + 2e- 2Cl-(aq)

2Fe2+ (aq) 2Fe3+

(aq) + 2e-

Step 3: Add half equations together and cancel out electrons

Cl2 (g) + 2Fe2+ (aq) 2Cl-

(aq) + 2Fe3+ (aq)

• In acid solution, dichromate(VI) ions (Cr2O72-) will oxidise sulphate(IV) ions

(SO32-) to sulphate(VI) (SO4

2-) ions and are themselves reduced to Cr3+.

Step 1: Write half equations

Reduction

Cr2O72- (aq) Cr3+

(aq)

Balance Cr2O72- (aq) 2Cr3+

(aq)

Cr2O72- (aq) + 14H+ (aq) 2Cr3+

(aq) + 7H2O (l)

Electrons +6 +3

Cr2O72- (aq) + 14H+ (aq) + 6e- 2Cr3+

(aq) + 7H2O (l)

• In acid solution, dichromate(VI) ions (Cr2O72-) will oxidise sulphate(IV) ions

(SO32-) to sulphate(VI) (SO4

2-) ions and are themselves reduced to Cr3+.

Step 1: Write half equations

Oxidation

SO32- (aq) SO4

2- (aq)

Balance SO32- (aq) + H2O (l) SO4

2- (aq) + H+ (aq)

Electrons +4 +6

SO32- (aq) + H2O (l) SO4

2- (aq) + 2H+ (aq) + 2 e-

• In acid solution, dichromate(VI) ions (Cr2O72-) will oxidise sulphate(IV) ions

(SO32-) to sulphate(VI) (SO4

2-) ions and are themselves reduced to Cr3+.

Step 2: Balance electrons

Cr2O72- (aq) + 14H+ (aq) + 6e- 2Cr3+

(aq) + 7H2O (l)

3SO32- (aq) + 3H2O (l) 3SO4

2- (aq) + 6H+ (aq) + 6 e-

Step 3: Add the half equations and cancel.

Cr2O72- (aq) + 14H+ (aq) + 6e- + 3SO3

2- (aq) + 3H2O (l)

2Cr3+ (aq) + 7H2O (l) + 3SO4

2- (aq) + 6H+ (aq) + 6 e-

Cr2O72- (aq) + 8H+ (aq) + 3SO3

2- (aq) 2Cr3+ (aq) + 4H2O (l) + 3SO4

2- (aq)

Let’s try it!

• Acidified manganite(VII) ions (MnO4-) can be reduced to Mn2+ by Fe2+

ions. The Fe2+ ions are oxidised to Fe3+. Write a balanced equation for this reaction.

• MnO4- (aq) + 8H+ (aq) + 5Fe2+

(aq) Mn2+ (aq) + 4H2O (l) + 5Fe3+

(aq)

14.2 Electrode Potentials and the Electrochemical SeriesLearning Objectives:1. Write half equations for reactions at electrodes.2. Describe what is meant by “standard electrode potential”.3. Describe how standard electrode potentials are measured.4. Calculate electrode potentials for an electrochemical cell.5. Draw a representation of an electrochemical cell.

Electrochemical Cells

• If two different metals are dipped into salt solutions containing their own ions and connected by a wire an electric current will flow.• The electrons will flow from the more

reactive metal (gives up electrons more easily to form + ions) to the less reactive metal.• Reduction happens at one electrode and

oxidation happens at the other.

Zn Cu

Zn2+ Cu2+

e-

Electrochemical Cells

• Zn is more reactive than Cu.

• Zn is oxidised (loses electrons so is a reducing agent).• Zn Zn2+ + 2e-

• Cu2+ ions in solution are reduced (gains electrons so is an oxidising agent)• Cu2+ + 2e- Cu

Zn Cu

Zn2+ Cu2+

e-

Potential Difference (Voltage)

• Zn acquires a positive charge.• Zn is said to gain a negative electrical

potential. This represents how easily electrons are lost.• However, this electrical potential cannot be

measured.

• But the potential difference between two electrodes can be measured using a voltmeter.

The Standard Hydrogen Electrode• The standard hydrogen electrode is

used as a “control” to compare metals.• Bubble H2 gas into a solution of H+ ions.• A platinum electrode is used to conduct

electricity. Platinum is used as it is a unreactive metal.• Standard conditions: H+ 1.00 mol/dm-3,

100 kPa, 298 K• The electrode potential is defined as

zero, so the potential difference of any half cell attached to is the electrode potential of that metal.

The Electrochemical Series• Electrode potentials can be put into

a table called the electrochemical series.• Usually listed with most negative at

the top (best reducing agents) • To least negative (positive) at the

bottom (best oxidising agents).• All reactions are shown as reduction

reactions.• If oxidation takes place, switch the

sign of the electrode potential.

Electrode Potentials Can Be Used to Calculate the Cell Potential

• Cell potential (E), also called electromotive force (e.m.f.), is the potential difference (voltage) between two half cells.• This can be calculated using the

electrode potentials.

Calculating Cell Potentials

1. Look up the electrode potentials for the two half cells.2. Write the reaction for the more negative electrode potential as an

oxidation reaction and switch the sign of the electrode potential.3. Add the two electrode potentials together.

Example: Zinc/Copper Electrochemical Cell• Connecting two half cells creates an

electrochemical cell.

What reactions are taking place?• Zn has a more negative electrode

potential, so is the stronger reducing agent, meaning it is oxidised.• Zn Zn2+ + 2e- E = +0.76 V• Copper is then reduced.• Cu2+ + 2e- Cu E = +0.34 V

Half Reaction Eθ (V)Zn2+ + 2e- Zn -0.76Cu2+ + 2e- Cu +0.34

Example: Zinc/Copper Electrochemical Cell

• Zn Zn2+ + 2e- Eθ = +0.76 V• Cu2+ + 2e- Cu Eθ = +0.34 V

• To find the cell potential, add the two electrode potentials together (remember to be careful of the signs!).

• Eθcell = (+0.76 V) + (+0.34 V) = +1.10 V

Half Reaction Eθ (V)Zn2+ + 2e- Zn -0.76Cu2+ + 2e- Cu +0.34

Drawing Electrochemical Cells

• To simplify, there is a standard shorthand way to represent electrochemical cells.

Example: Lithium Cell

• The two half-equations below show the reactions at the positive and negative electrodes of a lithium cell.

• Li+ (aq) + e- Li (s) Eθ = -3.04 V

• Li+ (aq) + MnO2 (s) + e- LiMnO2 (s) Eθ = -0.13 V

• A platinum rod is used as the positive electrode. Draw a representation of the cell.

• Li+ (aq) + e- Li (s) Eθ = -3.04 V

• Li+ (aq) + MnO2 (s) + e- LiMnO2 (s) Eθ = -0.13 V

• The top equation has the more negative electrode potential, so will go on the left side of the cell representation (oxidation will happen, so we flip it).• Li (s) Li+ (aq) + e-

Li (s) I Li+ (aq) This is the first half cell.• The second equation will go on the right side and remain as it is (reduction).• Li+ (aq) + MnO2 (s) + e- LiMnO2 (s)

Li (s) l Li+ (aq) II Li+ (aq) I MnO2 (s) , LiMnO2 (s)

Now add in the Pt electrode at the end.

Li (s) l Li+ (aq) II Li+ (aq) I MnO2 (s) , LiMnO2 (s) I Pt

14.3 Predicting Redox Reactions

Learning Objectives:1. Use electrode potentials to predict if a redox reaction will occur.

Predicting if a redox reaction will occur1. Write the two half equations for the redox reaction but

write them both as reduction reactions.2. Use the electrochemical series to find which has the more

negative electrode potential.3. Write the equation with the more negative electrode

potential as an oxidation reaction.4. Add the half equations together.5. The reaction is feasible in this direction, but not the other.

Example: Will zinc react with aqueous copper ions?1. Half equations as reductions

Zn2+ + 2e- ZnCu2+ + 2e- Cu

2. Find the more negative electrode potentialZinc E = -0.76 V Copper = +0.34 V

3. Write this as oxidationZn Zn2+ + 2e-

4. Add half equations togetherZn + Cu2+ Zn2+ + Cu

5. Is it feasible?Yes, the reaction is in the correct direction.