Electrode Potentials

know the IUPAC convention for writing half-equations for electrode reactions.

Know and be able to use the conventional representation of cells.

Know that standard electrode potential, E , refers to conditions of 298 K, 100 kPa and 1.00 mol dm−3 solution of ions.

Zn2+(aq) + 2 e– Zn(s)

Zn

Zn Zn2+ + 2 e-

oxidation

Cu2+ + 2 e- Cureduction

- electrode

anodeoxidation

+ electrodecathode

reductionelectron flow

At this electrode the metal loses

electrons and so is oxidised to metal

ions.

These electrons make the electrode

negative.

At this electrode the metal ions gain

electrons and so is reduced to metal

atoms.

As electrons are used up, this makes the electrode positive.

Cu

Standard Conditions

Concentration 1.0 mol dm-3 (ions involved in ½ equation)

Temperature 298 K

Pressure 100 kPa (if gases involved in ½ equation)

Current Zero (use high resistance voltmeter)

S tandard H ydrogen E lectrode

Emf = E = Eright - Eleft

H2 at 100 kPa

o

o

o

o

o

o

o

o

o

o

o

o

salt bridge

1.0 M H+(aq)

Pt

temperature= 298 K

1.0 M Cu2+(aq)

V

Cu

high resistancevoltmeter

E = Eright

H2 at 100 kPa

o

o

o

o

o

o

o

o

o

o

o

o

salt bridge

1.0 M H+(aq)

Pt

temperature= 298 K

1.0 M Cu2+(aq)

V

Cu

high resistancevoltmeter

Pt(s) | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)

These give a diagrammatic representation of what is happening in a cell.

• Place the cell with the more positive E° value on the RHS of the diagram.

Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS

Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS

ZINC IS IN CONTACT THE SOLUTIONS A SOLUTION OFWITH A SOLUTION ARE JOINED VIA A COPPER IONS IN OF ZINC IONS SALT BRIDGE TO CONTACT WITH COPPER

CELL DIAGRAMSCELL DIAGRAMS

Zn Zn2+ Cu2+ Cu

These give a diagrammatic representation of what is happening in a cell.

• Place the cell with the more positive E° value on the RHS of the diagram.

Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS

Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS

• Draw as shown… the cell reaction goes from left to right

• the zinc metal dissolves Zn(s) ——> Zn2+(aq) + 2e¯ OXIDATION

• copper is deposited Cu2+(aq) + 2e¯ ——> Cu(s) REDUCTION

• oxidation takes place at the anode

• reduction at the cathode

CELL DIAGRAMSCELL DIAGRAMS

Zn Zn2+ Cu2+ Cu

+_

CELL DIAGRAMSCELL DIAGRAMS

These give a diagrammatic representation of what is happening in a cell.

• Place the cell with the more positive E° value on the RHS of the diagram.

Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS

Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS

• Draw as shown… the electrons go round the external circuit from left to right

• electrons are released when zinc turns into zinc ions

• the electrons produced go round the external circuit to the copper

• electrons are picked up by copper ions and copper is deposited

Zn Zn2+ Cu2+ Cu

V

+_

CELL DIAGRAMSCELL DIAGRAMS

These give a diagrammatic representation of what is happening in a cell.

• Place the cell with the more positive E° value on the RHS of the diagram.

Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS

Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS

• Draw as shown… the cell voltage is E°(RHS) - E°(LHS) - it must be positive

cell voltage = +0.34V - (-0.76V) = +1.10V

Zn Zn2+ Cu2+ Cu

V

+_

Ni(s) | Ni2+(aq) || Sn4+(aq), Sn2+(aq) | Pt(s)

K(s) | K+(aq) || Mg2+(aq) | Mg(s)

ROOR

Standard electrode potentials E/V

F2(g) + 2 e- 2 F-(aq) + 2.87

MnO42-(aq) + 4 H+(aq) + 2 e- MnO2(s) + 2 H2O(l) + 1.55

MnO4-(aq) + 8 H+(aq) + 5 e- Mn2+(aq) + 4 H2O(l) + 1.51

Cl2(g) + 2 e- 2 Cl-(aq) + 1.36

Cr2O72-(aq) + 14 H+(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l) + 1.33

Br2(g) + 2 e- 2 Br-(aq) + 1.09

Ag+(aq) + e- Ag(s) + 0.80

Fe3+(aq) + e- Fe2+(aq) + 0.77

MnO4-(aq) + e- MnO4

2-(aq) + 0.56

I2(g) + 2 e- 2 I-(aq) + 0.54

Cu2+(aq) + 2 e- Cu(s) + 0.34

Hg2Cl2(aq) + 2 e- 2 Hg(l) + 2 CI-(aq) + 0.27

AgCl(s) + e- Ag(s) + Cl-(aq) + 0.22

2 H+(aq) + 2 e- H2(g) 0.00

Pb2+(aq) + 2 e- Pb(s) - 0.13

Sn2+(aq) + 2 e- Sn(s) - 0.14

V3+(aq) + e- V2+(aq) - 0.26

Ni2+(aq) + 2 e- Ni(s) - 0.25

Fe2+(aq) + 2 e- Fe(s) - 0.44

Zn2+(aq) + 2 e- Zn(s) - 0.76

Al3+(aq) + 3 e- Al(s) - 1.66

Mg2+(aq) + 2 e- Mg(s) - 2.36

Na+(aq) + e- Na(s) - 2.71

Ca2+(aq) + 2 e- Ca(s) - 2.87

K+(aq) + e- K(s) - 2.93

Increasingreducing

power

Increasingoxidising

power

GOLDEN RULE

The more +ve electrode gains electrons

(+ charge attracts electrons)

Electrodes with negative emf are better at releasing electrons (better reducing agents).

• A2.CHEM5.3.003

5.3 EXERCISE 2 - electrochemical cells

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q1

- 2.71 = Eright - 0

Eright = - 2.71 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q2

Emf = - 0.44 - 0.22

Emf = - 0.66 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q3

Emf = - 0.13 - (-0.76)

Emf = + 0.63 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q4

+1.02 = +1.36 - Eleft

Eleft = + 1.36 - 1.02 = +0.34 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q5

a) Emf = + 0.15 - (-0.25) = +0.40 Vb) Emf = + 0.80 - 0.54 = +0.26 Vc) Emf = + 1.07 - 1.36 = - 0.29 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q6

a) Eright = +2.00 - 2.38 = - 0.38 V

Ti3+(aq) + e- Ti2+(aq)

b) Eleft = -2.38 - 0.54 = - 2.92 V

K+(aq) + e- K(aq)c) Eright = - 3.19 + 0.27 = - 2.92 V Ti3+(aq) + e- Ti2+(aq)

ELECTRODE POTENTIALS – Q7

Emf = -0.76 - (-0.91) = +0.15 V

a) Cr(s) | Cr2+(aq) || Zn2+(aq) | Zn(s)

Emf = +0.77 - 0.34 = +0.43 V

b) Cu(s) |Cu2+(aq)|| Fe3+(aq),Fe2+(aq)| Pt(s)

Emf = +1.51 – 1.36 = +0.15 V

c) Pt(s) | Cl-(aq)| Cl2(g) || MnO4-(aq),H+(aq),Mn2+(aq)| Pt(s)