Download B9 – Respiration

B9 – Respiration(HL)
B.9.1  Compare aerobic and anaerobic respiration of glucose in terms of oxidation/reduction and
energy released.
[In aerobic respiration, glucose is converted into pyruvate, which, in the presence of oxygen, changes to
carbon dioxide and water. Overall, glucose undergoes oxidation and oxygen undergoes reduction. In
anaerobic respiration, pyruvate is converted to lactate in human beings, whereas yeast converts pyruvate to
ethanol and carbon dioxide. Redox equations should be used as appropriate.]
RESPIRATION
 Respiration is the metabolism of glucose(from carbohydrates) [and sometimes amino acids(from proteins)
and fatty acids(from fats)] to produce energy for the body. Respiration can take place aerobically(in the
presence of oxygen) or anaerobically(in the absence of oxygen).
A. AEROBIC(Oxidative) RESPIRATION
* Most of this energy is produced via aerobic respiration. Aerobic respiration takes place when the oxygen
supply is sufficient.
Activity:
a) Assign oxidation numbers to all atoms in the overall equation.
b) Determine the OA and the RA.
c) Extract the “foundation” of each ½ rxn.
d) Balance each ½ rxn.
Overall:
C6H12O6 + 6O2  6CO2 + 6H2O
ΔHc° = -2803 kJ mol-1
Oxidaton ½ rxn:
Reduction ½ rxn:
B. ANAEROBIC RESPIRATION
When oxygen supply is insufficient, (when engaged in high intensity exercise, for example),
energy can still be produced via anaerobic respiration.
In humans: glucose  pyruvic acid  lactic acid
In yeast: glucose  pyruvic acid  ethanol and CO2
Summary:
ANAEROBIC RESPIRATION
AEROBIC RESPIRATION
IN HUMANS
IN YEAST
glucose  lactic acid
glucose  ethanol + CO2
glucose + O2  CO2 + H2O
C6H12O6  2C3H6O3
C6H12O6  C2H5OH + CO2
C6H12O6 + 6O2  6CO2 + 6H2O
(fermentation)
-1
ΔH° = -77 kJ mol
ΔH° = -69 kJ mol-1
ΔHc° = -2803 kJ mol-1
Not a redox rxn. (*i.e. no Redox rxn.
Redox rxn.
Δ oxidation #’s !)
*Note the relative amounts of energy given off by each process.

 Pyruvic Acid
Carbohydrates  Glucose Glycolysis
i.e. C3H4O3
H+ + C3H3O3-
H+ + Pyruvate
[Draw pyruvic acid(propanoic acid with carbonyl on carbon 2]
GLUCOSE TO PYRUVATE(Anaerobic):
C6H12O6 + NAD+  2C3H3O3- + 2NADH + 4H+ + energy
(Assign oxidation numbers to atoms in glucose and pyruvate. Is glucose an OA or RA?
B.9.2  Outline the role of copper ions in electron transport and iron ions in oxygen transport.
[Cytochromes and hemoglobin are suitable examples.]
Review: Outline the four major properties of transition metals.
*1. Multiple oxidation states (application = electron transport/catalysis)
*2. Ability to from complex ions (application = oxygen transport)
*3. Catalysts
4. Coloured compounds
Electron Transport(e.g. cytochromes)
Transition metal ions help catalyze some of the redox reactions involved in aerobic respiration. The transition
metal ion is held inside a type of protein called a cytochrome.
Example: Cu2+ catalyzes the reduction of oxygen to water via a process known as electron transport. This
takes advantage of the transition metal’s ability to change readily between two oxidation states.
Step 1: Cu2+ accepts an electron(i.e. gets reduced)
i.e.
Cu2+ + e- (from oxidation pyruvic acid)  Cu+
Step 2:
Cu+ gives up its extra e-(i.e. gets oxidized) to reduce the oxygen to water.
i.e.
Oxidation:
Reduction:
4Cu+  Cu2+ + 4eO2 + 4H+ + 4e-  2H2O
Oxygen Transport(e.g. hemoglobin)
*Recall: Transition metals are able to form complex ions by binding to ligands (i.e. molecules or ions with a
non-bonding pair of electrons).
A hemoglobin molecule contains four heme groups. Each heme group contains a Fe2+ ion. Each Fe2+ ion can
accommodate one O2 ligand (…plus the 5 ligands that are already there). Therefore, one molecule of
hemoglobin can transport a maximum of four oxygen molecules/ligands.
Heme Group
Heme group with O2 ligand