In a standard hydrogen electrode, why does increasing the partial pressure of H2 gas reduce the standard potential (E) of the cell?
Nevermind I figured it out
To understand why increasing the partial pressure of H2 gas reduces the standard potential (E) of a standard hydrogen electrode, it is important to understand the concept of the Nernst equation and the relationship between electrode potential and reactant concentrations.
The Nernst equation is used to calculate the electrode potential (E) of an electrochemical cell based on the concentrations (or partial pressures) of the reactants involved. It is given as:
E = E° - (RT/nF) * ln(Q)
Where:
- E is the electrode potential
- E° is the standard electrode potential
- R is the gas constant
- T is the temperature
- n is the number of electrons transferred in the reaction
- F is the Faraday constant
- Q is the reaction quotient, calculated using the concentrations (or partial pressures) of the reactants.
For the standard hydrogen electrode, the standard electrode potential (E°) is defined as zero volts. This means that the standard potential of the hydrogen electrode is considered the reference point against which the potentials of other electrodes are measured.
In the case of the standard hydrogen electrode, the half-reaction involved is the reduction of H+ ions to form H2 gas:
2H+ + 2e- -> H2
As per the Nernst equation, an increase in the partial pressure of H2 gas will result in a decrease in Q, the reaction quotient. Since ln(Q) is negative when Q is less than 1, a decrease in Q will cause the term (RT/nF) * ln(Q) to become more positive. Consequently, this positive value subtracted from the zero volts standard electrode potential will result in a reduction in the overall cell potential. In simpler terms, increasing the partial pressure of H2 gas favors the forward reaction and lowers the electrode potential.
In summary, increasing the partial pressure of H2 gas reduces the standard potential of the standard hydrogen electrode because it affects the reaction quotient, leading to a decrease in the potential according to the Nernst equation.