Give the cell diagram notation of the electrochemical cell that could be used to determine experimentally the dissociation constant (Kw) of water. The standard reduction potential for
O2 + H20 + 4 e- ----> 4OH- is 0.40 V
For other possible half-reactions, see your textbook.
Calculate E° for this cell.
Also, calculate the dissociation constant at 298K.
To determine the dissociation constant (Kw) of water experimentally, we can use an electrochemical cell. The cell diagram notation for this cell would be:
Pt | H2 (g) | H+ (aq) || OH- (aq) | O2 (g) | Pt
In this cell diagram, Pt represents a platinum electrode that is inert and does not participate in the reaction. H2 (g) represents a hydrogen gas electrode, H+ (aq) represents a solution of acid, OH- (aq) represents a solution of alkali, and O2 (g) represents a gas electrode.
The standard reduction potential for the half-reaction O2 + H2O + 4e- -> 4OH- is given as 0.40 V.
To calculate E° for this cell, we need to consider the half-reactions occurring at each electrode:
At the hydrogen gas electrode (anode):
H2 (g) -> 2H+ (aq) + 2e-
At the gas electrode (cathode):
O2 (g) + 2H2O (l) + 4e- -> 4OH- (aq)
The overall cell reaction is the sum of these two half-reactions. Since the hydrogen gas electrode is the anode, it will have a negative sign, while the gas electrode is the cathode and will have a positive sign:
Overall cell reaction:
H2 (g) + O2 (g) + 2H2O (l) -> 2H+ (aq) + 4OH- (aq)
Now we can calculate the standard cell potential, E°, by summing the standard reduction potentials for each half-reaction:
E° = E°(cathode) - E°(anode)
= 0.40 V - 0 V (since the half-reaction at the hydrogen gas electrode has no potential)
Therefore, E° for this cell is 0.40 V.
To calculate the dissociation constant (Kw) at 298K, we can use the Nernst equation:
E = E° - (RT/nF) * ln(Q)
Where:
E = Cell potential at a particular condition
E° = Standard cell potential
R = Gas constant (8.314 J/mol*K)
T = Temperature in Kelvin (298K)
n = Number of electrons transferred
F = Faraday's constant (96,485 C/mol)
Q = Reaction quotient
In the case of water dissociation, the reaction quotient Q is given by:
Q = [OH-]^4 / [H+]^2
Since Kw is defined as [OH-][H+], we can rewrite Q as:
Q = [OH-]^2 / [H+]^2
Plugging these values into the Nernst equation:
E = E° - (RT/nF) * ln([OH-]^2 / [H+]^2)
At 298K, substituting the values into the equation:
E = 0.40 V - (8.314 J/mol*K * 298K / (4 * 96,485 C/mol)) * ln([OH-]^2 / [H+]^2)
Now, we need to rearrange the equation and solve for Kw:
Kw = [OH-]^2 / [H+]^2 = 10^(2 * E / (0.05916 V))
By substituting the value of E, which is 0.40 V, into the equation:
Kw = 10^(2 * 0.40 V / (0.05916 V))
Solving this equation will give you the dissociation constant (Kw) at 298K.