If E Cu+2 had been chosen as the standard reference electrode and had been assigned a potential of 0.00 V, what would the reduction potential of the hydrogen electrode be relative to it?
Ag.............0.8.........?
Cu ...........+0.34.........0
H..............0..........-0.34
Ni...........-0.23..........?
You just move everything down the page from Cu to h so you reduce everything by 0.34.
To determine the reduction potential of the hydrogen electrode relative to E Cu+2, we can use the Nernst equation:
Ecell = Eo cell - (RT / nF) * ln(Q)
In this case, the standard reduction potential of E Cu+2 is 0.00 V. Since it is the reference electrode, its Eo value is considered zero.
Given that the hydrogen electrode is the other half-cell, and we want to find its reduction potential (Eo H2), we can rewrite the equation as:
Ecell = Eo H2 - 0 - (RT / nF) * ln(Q)
Since we are comparing it to the E Cu+2 reference, the Ecell value is also zero.
0 = Eo H2 - 0 - (RT / nF) * ln(Q)
Simplifying further:
0 = Eo H2 - (RT / nF) * ln(Q)
Now, we need to consider the values for the gas constant (R), temperature (T), the Faraday constant (F), and the reaction quotient (Q) for the hydrogen electrode.
The Faraday constant, F, is approximately 96485 C/mol.
The ideal gas constant, R, is approximately 8.314 J/(mol·K).
The reaction quotient, Q, represents the concentration of the species involved in the half-cell reaction, which is the concentration of hydrogen ions (H+) divided by the hydrogen gas (H2) concentration. Since both are 1 M at standard conditions, Q is equal to 1.
Substituting the values into the equation:
0 = Eo H2 - (8.314 * T / (n * 96485)) * ln(1)
Since ln(1) equals 0, the equation becomes:
0 = Eo H2 - 0
Thus, the reduction potential of the hydrogen electrode relative to E Cu+2 would be zero volts, indicating that both are at the same potential.
To determine the reduction potential of the hydrogen electrode relative to the chosen standard reference electrode, you'll need to use the Nernst equation. The Nernst equation relates the standard reduction potential of a half-cell reaction to the actual reduction potential under nonstandard conditions.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
- E is the reduction potential under nonstandard conditions
- E° is the standard reduction potential
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- n is the number of moles of electrons transferred in the balanced half-reaction
- F is Faraday's constant (96485 C/mol)
- ln(Q) is the natural logarithm of the reaction quotient
In this case, the reduction potential of the hydrogen electrode is what we're trying to find, while E Cu+2 (0.00 V) is the standard reduction potential of the chosen reference electrode.
The balanced half-reaction for the hydrogen electrode is:
2H+ + 2e- → H2
The number of moles of electrons transferred (n) for this balanced half-reaction is 2.
Plugging in the known values:
E = 0.00 V - (8.314 J/mol·K * Temperature/K / (2 * 96485 C/mol)) * ln(Q)
Since we're comparing the reduction potential of the hydrogen electrode to E Cu+2, the reaction quotient (Q) will be equal to 1, as the concentration of Cu+2 is 1 M in a standard hydrogen electrode.
Simplifying the equation further:
E = 0.00 V - (8.314 J/mol·K * Temperature/K / (2 * 96485 C/mol)) * ln(1)
Since ln(1) is equal to 0:
E = 0.00 V - (8.314 J/mol·K * Temperature/K / (2 * 96485 C/mol)) * 0
Therefore, the reduction potential of the hydrogen electrode relative to E Cu+2 when E Cu+2 is assigned a potential of 0.00 V is also 0.00 V. This means that the hydrogen electrode is at the same potential as the chosen standard reference electrode.