Lecture 22

Midterm 2 Statistics
Mean: 37/60 (61.5%)
High score: 60
Minimum score: 12

C+

40% (final) + 20% (homework) of the grade remains to be determined! Standard Reduction Potentials
Standard reduction potential (E0) is the potential associated with a reduction reaction at an electrode when all solutes are
1 M and all gases are at 1 atm (standard conditions)
Standard Hydrogen Electrode (SHE) = accepted reference point

Reduction Reaction
2e- + 2H+ (1 M)

H2 (1 atm)

Standard potential assigned to E0 = 0 V
All other standard reduction potentials are measured relative to the SHE reference

Standard Reduction Potentials
• Electrode potentials are always measured relative to each other
• The absolute potential scale is set by picking one electrode as an arbitrary reference point
• H+ (1 M) | H2 (1 atm) | Pt (s) electrode is assigned a potential of 0 volts
• All other potentials are measured relative to it

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Anode (oxidation):

Zn (s)

Cathode (reduction): 2e- + 2H+ (1 M)
Zn (s) + 2H+ (1 M)

Zn2+ (1 M) + 2eH2 (1 atm)
Zn2+ + H2 (1 atm)

Standard Reduction Potentials
0
Ecell = 0.76 V (measured)

Must be cathode because electrons flow towards it
(experimentally determined)
0
Ecell definition:
0
0
0
Ecell = Ecathode - Eanode

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
0
0
0
Ecell = EH +/H2 - EZn2+/Zn

0
0.76 V = 0 - EZn2+/Zn

0
EZn 2+/Zn = -0.76 V

Zn2+ (1 M) + 2e-

Zn

E0 = -0.76 V

Standard Reduction Potentials
0
Ecell = 0.34 V (measured)

Must be cathode because electrons flow towards it
(experimentally determined)
0
0
0
Ecell = Ecathode - Eanode
0
0
0
Ecell = ECu 2+/Cu – EH +/H2
0
0.34 = ECu2+ /Cu - 0
0
ECu 2+ /Cu = 0.34 V

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode (oxidation):

H2 (1 atm)

Cathode (reduction): 2e- + Cu2+ (1 M)
H2 (1 atm) + Cu2+ (1 M)

2H+ (1 M) + 2eCu (s)

Cu (s) + 2H+ (1 M)

• Standard reduction potentials, E0
• The more positive E0 the greater the tendency for the substance to be reduced
• The sign of E0 changes when the reaction is reversed from reduction to oxidation (in a cell with SHE being the other electrode)
• Changing the stoichiometric coefficients of a half-reaction does not change the value of E0

EXAMPLE: What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
More positive of the two

Which half-reaction is reductive?
Cd2+ (aq) + 2e-

Cd (s) E0 = -0.40 V

Cd2+ will get reduced to Cd

Cr3+ (aq) + 3e-

Cr (s)

Cr will get oxidized to Cr3+

Anode (oxidation):

E0 = -0.74 V

Cr3+ (1 M) + 3e- x 2

Cr (s)

Cathode (reduction): 2e- + Cd2+ (1 M)
2Cr (s) + 3Cd2+ (1 M)

Cd (s) x 3

3Cd (s) + 2Cr3+ (1 M)

0
0
0
Ecell = Ecathode - Eanode
0
Ecell = -0.40 V – (-0.74 V)
0
Ecell = 0.34 V (positive = spontaneous)

Electrochemistry and Equilibrium
Reduction:

0
E An + ! A

An+ + n e- → A

E

Oxidation:

B → Bn+ + n e-

Overall:

0
B n+ ! B

0
0
0
Ecell = E An+ ! A " EBn+ ! B

An+ + B → A + Bn+

Under standard conditions:

ΔG0 = -nFEcell0 (electrical work per

one mole of the reaction)

n = number of moles of electrons in reaction
J
F = 96,500
= 96,500 C/mol (charge of
V • mol one mole of electrons)
ΔG0 = -RT ln Keq
0
Ecell =
At T = 298 K:

0
= -nFEcell