Consider the cell: (Pt) H2/H+ || (Pt) H+/H2. In the anode half-cell, hydrogen gas at 1.0 atm is bubbled over a platinum electrode dipping into a solution that has a pH of 7.0. The other half-cell is identical to the first except that the solution around the platinum electrode has a pH of 0.0. What is the cell voltage?
For the anode written as a reduction:
E = Eo - (0.0592/n)log(pH2/H^+)
Eo = 0, and H^+ = 1E-7, E then written as an oxidation will be the negative of that.
E for the cathode written as a reduction:
same equation with different numbers.
Then E cell = Eoxdn + Eredn
I think the answer is approximately 0.50v.
nevermind i got it.. the answer is -.414
thanks though
right. I dropped the - sign when I typed it in.
To determine the cell voltage of the given cell, we need to calculate the potential difference between the anode and cathode half-cells and then combine them.
The half-cell reactions can be written as follows:
Anode half-cell: H2(g) -> 2H+(aq) + 2e-
Cathode half-cell: 2H+(aq) + 2e- -> H2(g)
Since the reactions involve the same species, H2(g), on both sides, the standard hydrogen electrode (SHE) potential can be used as a reference. The SHE potential is considered to be 0.00 V.
To calculate the cell voltage, we need to determine the difference in potential between the anode and cathode half-cells. This difference is due to the difference in pH of the solutions around the platinum electrodes.
The Nernst equation relates the standard electrode potential to the actual cell potential in nonstandard conditions, taking into account the concentrations of species involved. It can be written as:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
Ecell = Cell potential
E°cell = Standard cell potential
n = Number of moles of electrons transferred in the balanced half-cell reactions
Q = Reaction quotient
In this case, the reaction quotient (Q) for both half-cells is simply 1, as the concentrations of all reactants are 1M. Therefore, the Nernst equation simplifies to:
Ecell = E°cell - (0.0592/n) * log(1)
Since both half-reactions involve 2 moles of electrons, n = 2. Plugging in the values, the equation becomes:
Ecell = E°cell - (0.0592/2) * log(1)
Ecell = E°cell - (0.0296) * log(1)
Given that the pH for the anode half-cell is 7.0 and for the cathode half-cell is 0.0, we can calculate the E°cell for each half-reaction using the Nernst equation at these pH values:
For anode half-cell:
E°anode = 0.00 V - (0.0592/2) * (7 - 0) = 0.00 V - 0.1776 V = -0.1776 V
For cathode half-cell:
E°cathode = 0.00 V - (0.0592/2) * (0 - 7) = 0.00 V + 0.2064 V = 0.2064 V
Now, we can calculate the cell voltage by subtracting the anode potential from the cathode potential:
Cell voltage = E°cathode - E°anode
Cell voltage = 0.2064 V - (-0.1776 V)
Cell voltage = 0.2064 V + 0.1776 V
Cell voltage = 0.384 V
Therefore, the cell voltage for the given cell is 0.384 V.