for this equation
Fe(CN)6 4- + Fe 3+ --> Fe(CN)6 3- + Fe 2+
at 310 K and pH 2
using the equation
E = Eo + 2.303 RT/nF log K - 2.303 pH RTh/nF
what is h ?
Isn't h the natural log of the reaction quotient?
what so for the last part of the equation -2.303 RT/nF ln pH?
To solve for "h" in the equation E = Eo + 2.303 RT/nF log K - 2.303 pH RTh/nF, we need to understand each term in the equation:
- E: This represents the cell potential.
- Eo: This refers to the standard cell potential, which is the cell potential under standard conditions.
- R: This is the ideal gas constant (8.314 J/(mol·K)).
- T: This is the temperature in Kelvin (K).
- n: This represents the number of electrons involved in the balanced half-reaction.
- F: This is the Faraday constant (96,485 C/mol).
- log K: This refers to the logarithm of the equilibrium constant (K) for the given reaction.
- pH: This represents the pH value.
- Th: This is the thickness of the electrochemical cell.
However, in the given equation E = Eo + 2.303 RT/nF log K - 2.303 pH RTh/nF, there seems to be a mistake with the term "h." It should actually be "Th," representing the thickness of the electrochemical cell.
Thus, the value of "h" is not applicable to the equation provided. It should be corrected to "Th" instead.