Study Guide
Redox Reactions and Electrochemistry
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University:
Kentucky State University -
Course:
CHE 101 | General Chemistry I Academic year:
2022
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Views:
250
Pages:
17
Author:
penuchelnu2
Redox Reactions and
Electrochemistry
CHEM-102
1 Oxidation–Reduction Reactions
Involves two processes: Oxidation and Reduction
Oxidation and reduction always occur together
Electrochemical processes:
are oxidation-reduction reactions in which:
➢ the energy released by a spontaneous reaction
is converted to electricity or
➢ electrical energy is used to cause a nonspontaneous
reaction to occur
CHEM-102
2 Oxidation = Loss of electrons / Increase of Oxidation No.
Na ⎯→ Na+ + e–
Oxid. Half-Reaction
Reduction = Gain of electrons / Decrease of Oxidation No.
Cl2 + 2e– ⎯→ 2Cl–
Red. Half-Reaction
Net reaction:
2Na + Cl2 ⎯→ 2Na+ + 2Cl–
0
0
2Mg (s) + O2 (g)
2Mg
O2 + 4eCHEM-102
2Mg2+ + 4e-
2O2-
2+ 2-
2MgO (s)
Oxidation
Reduction
3 In the following reaction, which species is the oxidizing
agent and which gets reduced?
2H+(aq) + Mn(s) ⎯→ Mn2+(aq) + H2(g)
A. H+ is the oxidizing agent and Mn gets reduced
B. H+ is the oxidizing agent and H+ gets reduced
C. Mn is the oxidizing agent and H+ gets reduced
D. Mn is the oxidizing agent and Mn gets reduced
E. Mn is the oxidizing agent and Mn2+ gets reduced
CHEM-102
4 Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation
number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to
the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In H2O2
and O22- it is –1.
CHEM-102
5 4. The oxidation number of hydrogen is +1 except when
it is bonded to metals in binary compounds. In these
cases, its oxidation number is –1.
5. Group IA metals are +1, IIA metals are +2 and fluorine is
always –1.
6. The sum of the oxidation numbers of all the atoms in a
molecule or ion is equal to the charge on the
molecule or ion.
CHEM-102
6 Galvanic Cells
anode
oxidation
cathode
reduction
spontaneous
redox reaction
CHEM-102
7 Galvanic Cells
0
Ecell
(+) Spontaneous
The difference in electrical
potential between the anode and
cathode is called:
• cell voltage
• electromotive force (emf)
• cell potential
Zn (s) + Cu2+ (aq)
Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M and [Zn2+] = 1 M
Cell Diagram
phase boundary
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
(-)
CHEM-102
salt bridge
(+)
8 Cell Diagram
Zn (s) + Cu2+ (aq)
Cu (s) + Zn2+ (aq)
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
CHEM-102
Anode
(-)
Cathode
(+)
Oxidation
Reduction
9 Standard Reduction Potentials
Standard reduction potential (E0) is the voltage associated
with a reduction reaction at an electrode when all solutes
are 1 M and all gases are at 1 atm.
Reduction Reaction
2e- + 2H+ (1 M)
H2 (1 atm)
E0 = 0 V
Standard hydrogen electrode (SHE)
CHEM-102
10 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)
CHEM-102
Zn2+ (1 M) + 2eH2 (1 atm)
Zn2+ (1 M) + H2 (1 atm)
11 0 = 0.76 V
Ecell
0 )
Standard emf (Ecell
0
0 = E0
Ecell
cathode - Eanode
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
0 = E 0 + - E 0 2+
Ecell
H /H2
Zn /Zn
0 2+
0.76 V = 0 - EZn
/Zn
0 2+
EZn
/Zn = -0.76 V
Zn2+ (1 M) + 2eCHEM-102
Zn
E0 = -0.76 V
12 0 = 0.34 V
Ecell
0
0 = E0
Ecell
cathode - Eanode
0 = E 0 2+
0
Ecell
Cu /Cu – EH +/H 2
0 2+
0.34 = ECu
/Cu - 0
0 2+
ECu
/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)
CHEM-102
2H+ (1 M) + 2eCu (s)
Cu (s) + 2H+ (1 M)
13 CHEM-102
•
The more positive E0 the
greater the tendency for the
substance to be reduced
•
The half-cell reactions are
reversible
•
The sign of E0 changes
when the reaction is
reversed
•
Changing the stoichiometric
coefficients of a half-cell
reaction does not change
the value of E0
14 Example 1:
A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2
solution and a Ag electrode in a 1.0 M AgNO3 solution.
Calculate the standard emf of this cell at 25°C.
Solution:
Ag+(1.0 M) + eMg2+(1.0 M) + 2e-
Ag(s)
Mg(s)
E° = 0.80 V
E° = -2.37 V
Anode (oxidation):
Mg(s)
Mg2+(1.0 M) + 2eCathode (reduction):
2Ag+(1.0 M) + 2e2Ag(s)
_________________________________________________
Overall:
Mg(s) + 2Ag+(1.0 M)
Mg2+(1.0 M) + 2Ag(s)
E°cell = E°cathode - E°anode
= E°Ag+/Ag - E°Mg2+/Mg
= 0.80 V - (-2.37 V)
= 3.17 V
CHEM-102
15 Example 2:
Predict what will happen if molecular bromine (Br2) is added to
a solution containing NaCl and NaI at 25°C. Assume all species
are in their standard states.
Solution:
Cl2(1 atm) + 2eBr2(l) + 2eI2(s) + 2e-
2Cl-(1 M)
2Br-(1 M)
2I-(1 M)
E° = 1.36 V
E° = 1.07 V
E° = 0.53 V
Br2 will oxidize I- but will not oxidize Cl-. Therefore,
Oxidation:
2I-(1 M)
I2(s) + 2eReduction:
Br2(l) + 2e2Br-(1 M)
Overall:
CHEM-102
2I-(1 M) + Br2(l)
I2(s) + 2Br-(1 M)
We can confirm our conclusion by calculating E°cell. Try it.
16 Cl2(1 atm) + 2eBr2(l) + 2eI2(s) + 2e-
CHEM-102
?
2Br-(1 M) + I2(l)
?
2Br-(1 M) + Cl2(l)
2Cl-(1 M)
2Br-(1 M)
2I-(1 M)
E° = 1.36 V
E° = 1.07 V
E° = 0.53 V
Br2(s) + 2I-(1 M)
X
Br2(s) + 2Cl-(1 M)
17
Redox Reactions and Electrochemistry
of 17
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