Lecture 5 - Stoichiometry and Bacterial Energetics II

Question 1

What is the oxidation state of N in NO3-?

Answer 1

1. Let x = O.S.
2. x(1) + (-1)(3) = -1
3. x = -2

Question 2

Construct the half reaction for the reduction of nitrite (NO2-) to ammonia (NH3).

Answer 2

1. O.S. N in NO2-
   x(1)+ (-2)(2) = -1
   x = 3
2. O.S. N in NH3
   x(1) + (1)(3) = 0
   x = -3

• 6 electrons transferred*
  3. NO2- + 6e- –> NH3

4. Balance O with H2O
   NO2- + 6e- –> NH3 + 2H2O
5. Balance H with H+
   NO2- + 6e- + 7H+ –> NH3 + 2H2O

Question 3

What is the alternative approach for constructing the half reaction? Use nitrite (NO2-) to ammonia (NH3) as an example.

Answer 3

1. Balance N
   NO2- –> NH3
2. Balance O with H2O
   NO2- –> NH3 + 2H2O
3. Balance H with H+
   NO2- + 7H+ –> NH3 + 2H2O
4. Balance the charge with electrons
   NO2- + 6e- + 7H+ –> NH3 + 2H2O

Question 4

Re

Answer 4

A reaction for energy production

To obtain Re we will combine two half reactions

• -> Ra - electron acceptor half reaction
• -> Rd - electron donor half reaction

Re = Ra - Rd

Question 5

Rs

Answer 5

A reaction for cellular synthesis

To obtain Rs we will combine two half reactions

• -> Rc - electron synthesis half reaction
• -> Rd - electron donor half reaction

Rs = Rc - Rd

Question 6

Given fe and fs, what is the overall equation R to describe bacterial cell growth?

Answer 6

Re = fe(Re) + fs(Rs); fe + fs = 1

shortcut: Re = fe(Ra) + fs(Rc) - Rd

• -> fe is the fraction of electrons goin to the electron acceptor for energy
• -> fs is the fraction of electrons used to synthesize cell