Energy balance of half-reactions and standard electrode potentials

　Consider the energy balance of the half-reaction equation, taking into account the energy possessed by the electrons.

　By the way, where there is an electric charge (such as an electron), an electric field is generated, and that charge has energy corresponding to the potential of the electric field.

The complicated explanation is not important, so remember the following.

"A quantity of charge q(C) at a certain potential E(V or J/C) has an energy (J) of q×E."

"The absolute value of the charge (C) of one mole of electrons (e-) is 96485 (C/mol), which is called Faraday constant (F)."

This is also complicated, so be sure to memorize only the following equation.

　Faraday constant is the absolute value of the amount of charge that one mole of electrons has. Since electrons have a negative charge, they have negative energy. Therefore, in the above equation, the potential is multiplied by the Faraday constant and the negative is attached to it.

Example of finding the standard electrode potential for a half reaction

　Let E_①(V) be the potential generated in the electron 4e^- that is transferred in the half reaction ① of organic matter oxidation.

　The unit of volt (V) can be converted to (J/C), where J (joule) is the unit of energy and C (coulomb) is the unit of charge. The absolute value of the amount of charge that one mole of electrons has is given by Faraday constant (F) 96485 (C/mol).

　Therefore, the energy possessed by the electron 4e^- in the half-reaction equation ① is -4・F・E_①(J).

　The standard Gibbs energy of formation for each substance, including the energy held by this electron, is noted below the half-reaction equation, and the equation is solved to balance the energies on the left and right sides of the half-reaction.

                          （original form）               （productform）

①                        CO₂  +   4H⁺   +    4e-   =    CH₂O    +     H₂O

G_f⁰ (kJ/mol)        －385        0                        －129.7      －237.18

electric energy(J)                       －4FE_①

Equation that balances the total energy of the left and right sides：

－385・10³　－4FE_①    =   －129.7・10³ －237.18・10³

Find the unknown number E_①.

E_①    =  (－129.7・10³ －237.18・10³＋ 385・10³)／(－4・96485) = －0.05 (V)

　This means that the total difference (product form - original form) between the standard Gibbs energy of formation of the left side (original form) and the right side (product form) is given to the four electrons. In a redox reaction, the energy given to these electrons is used to recombine chemical bonds.

　The potential calculated from that energy (E_①) is called the standard electrode potential for that half-reaction. Since the standard Gibbs energy of formation is unique to a substance, each half-reaction equation has a unique standard electrode potential.

In oxygen respiration,

Find the standard electrode potential of the half-reaction equation ② at which oxygen molecules are reduced.

Oxygen respiration (oxygen reduction)

②          　　　　　 O₂ 　+ 　4H⁺ 　　+　4e^- 　　= 　　 2H₂O    

G_f⁰ (kJ/mol)        　　0         0                                 2（－237.18）

Electric energy (J)                             －4FE_②

Energy balance：－4FE_②　＝ 2（－237.18・10³）

E_② =  2（－237.18・10³）／ (－4・96485) = 1.23 (V)                                             

Sulfate respiration (sulfate reduction)

④                      SO₄^2-   +   10H⁺   +   8e^-   =  H₂S   +   4H₂O

G_f⁰ (kJ/mol) 　　－744.5                  0                      －27.78    4(－237.18)

Electric energy (J)                             －8FE_④

Energy balance：－744.5・10³－8FE_④　＝　－27.78・10³－4・237.18・10³

E_④ =  （－27.78・10³－4・237.18・10³＋744.5・10³）／ (－8・96485) = 0.30 (V)

※Remember to multiply by 1000 because of the kilo (k) unit of Gibbs energy.