Determine the unknown quantity in the following cell at 298.15 K.
Pt(s) | H2(g, 1.0 bar)| H+(pH = ?) || Cl -(aq, 1.0 mol·L-1) |AgCl(s), Ag(s), ΔE = +0.36 V.
(DrBob) I still wasn't able to figure out the other one--I posted again after that one, but I know there are a lot of posts. This is a similar question, but I still can't figure out how to set up the equations properly. I know I need to find the concentration of H and convert it to pH (which I know how to do). I have no idea how to incorporate the pressure into it (I know here it is 1 so it may not affect my equation...but in my other problem it is 2 bar and so I don't know what to do with it).
Thanks!!!
To determine the unknown quantity in this cell at 298.15 K, we need to set up the appropriate half-cell reactions and use the Nernst equation.
First, let's analyze the given cell diagram:
Pt(s) | H2(g, 1.0 bar) | H+(pH = ?) || Cl-(aq, 1.0 mol·L-1) | AgCl(s), Ag(s), ΔE = +0.36 V.
From this cell diagram, we can identify the following:
- The left-hand side of the cell (Pt(s) | H2(g, 1.0 bar) | H+(pH = ?)) represents the reduction half-cell.
- The right-hand side of the cell (Cl-(aq, 1.0 mol·L-1) | AgCl(s), Ag(s)) represents the oxidation half-cell.
- The ΔE = +0.36 V gives us the standard cell potential, which we'll use later.
Now, let's set up the reduction and oxidation half-cell reactions:
Reduction half-cell: H2(g, 1.0 bar) | H+(pH = ?)
Oxidation half-cell: Cl-(aq, 1.0 mol·L-1) | AgCl(s), Ag(s)
The balanced half-cell reactions are:
Reduction: 2H+ + 2e- -> H2(g)
Oxidation: 2AgCl(s) -> 2Ag(s) + Cl2(g) + 2e-
Next, we need to write the overall cell reaction by adding these two half-cell reactions:
Overall cell reaction: 2H+(aq) + 2AgCl(s) -> H2(g) + 2Ag(s) + Cl2(g)
Now, we can use the Nernst equation to determine the pH at 298.15 K:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell = Cell potential
E°cell = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin (298.15 K)
n = Number of electrons transferred in the balanced cell reaction (in this case, 2)
F = Faraday's constant (96485 C/mol)
Q = Reaction quotient
Since the standard cell potential (E°cell) is given as +0.36 V, we can substitute the values into the Nernst equation and rearrange to solve for pH, as follows:
Ecell = E°cell - (RT/nF) * ln(Q)
0.36 V = 0.36 V - (8.314 J/(mol·K) * 298.15 K / (2 * 96485 C/mol) * ln(Q)
To isolate ln(Q), we can cancel out the standard cell potential terms:
0 = - (8.314 J/(mol·K) * 298.15 K / (2 * 96485 C/mol) * ln(Q))
This equation simplifies to:
ln(Q) = 0
For any non-zero value of Q, the natural logarithm cannot be equal to zero. Therefore, the only solution is ln(Q) = 0.
If ln(Q) = 0, it means Q = e^0 = 1, which leads to a neutral pH.
Hence, the unknown quantity in the given cell at 298.15 K is pH = 7 (neutral).