To what pH should you adjust a standard hydrogen electrode to get an electrode potential of -0.122 V ? (Assume that the partial pressure of hydrogen gas remains at 1 atm.)
To answer this question, we need to understand the concept of Nernst equation, which relates the electrode potential of a half-cell to the concentration (or activity) of the involved species.
The Nernst equation for a half-cell reaction can be written as:
E = E° - (RT / nF) * ln(Q)
where:
E = electrode potential of the half-cell
E° = standard electrode potential
R = gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
n = number of electrons transferred in the half-reaction
F = Faraday's constant (96485 C/mol)
Q = reaction quotient
In this case, we are dealing with a standard hydrogen electrode (SHE), which has a standard electrode potential of 0 V. We need to find the pH at which the electrode potential is -0.122 V.
Since the SHE involves the reduction of H+ ions to form H2 gas, the half-reaction can be written as:
2H+ + 2e- -> H2
Now, let's substitute the known values into the Nernst equation and solve for the required pH.
E = -0.122 V (given)
E° = 0 V (for standard hydrogen electrode)
R = 8.314 J/(mol·K)
T = temperature (which is not given, assume room temperature: 298 K)
n = 2 (since 2 electrons are transferred in the half-reaction)
F = 96485 C/mol
Q = activity of the hydrogen ion (H+) * activity of hydrogen gas (H2 gas pressure)
Since the partial pressure of the hydrogen gas is given as 1 atm, the activity of the H2 gas is equal to 1.
Substituting the values into the equation:
-0.122 V = 0 V - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol) * ln(Q)
Simplifying:
-0.122 V = -0.0592 V * ln(Q)
Now, we need to solve for Q by rearranging the equation:
ln(Q) = -0.122 V / -0.0592 V
ln(Q) ≈ 2.06
Now, we can find Q by taking the inverse exponential (antilog) of both sides:
Q ≈ e^2.06 ≈ 7.84
In the case of the pH, it represents the negative logarithm (base 10) of the hydrogen ion activity (H+):
pH = -log[H+]
Since Q represents the activity of H+, we can write:
Q = [H+]
Now, let's take the negative logarithm (base 10) of both sides:
-pH = log(Q)
pH = -log(Q)
Substituting the value of Q:
pH = -log(7.84)
pH ≈ 0.11
Therefore, to adjust the standard hydrogen electrode to get an electrode potential of -0.122 V, you should adjust the pH to approximately 0.11.