If the voltage of the cell is -0.1288V at 25Celsius when [Pb2+]=0.1403M and the pressure of H2=1atm, what is the pH
Pb(s)+2H+(aq)<-> Pb2+(aq) +H2(g) E0 = 0.1260V
Ecell = Eocell - (0.05916/2)log(Pb^2+)*PH2/(H^+)^2
Substitute and solve for H^+, then convert to pH.
Thank you DrBob222
To determine the pH of the solution, we need to consider the Nernst equation, which relates the cell potential to the concentrations of the reacting species.
The Nernst equation is given as follows:
E = E0 - (RT/nF) * ln(Q)
Where:
E is the cell potential,
E0 is the standard cell potential,
R is the gas constant (8.314 J/(mol·K)),
T is the temperature in Kelvin,
n is the number of electrons involved in the reaction (in this case, 2),
F is the Faraday constant (96485 C/mol),
ln is the natural logarithm,
and Q is the reaction quotient.
First, let's calculate the reaction quotient Q using the given concentrations:
Q = [Pb2+]/[H+]^2
Q = 0.1403 / [H+]^2
To proceed, we need to determine the value of [H+].
From the cell reaction equation, we can infer that 2 moles of H+ ions are produced for every 1 mole of Pb2+ ions. At equilibrium, the concentration of H+ ions is related to the concentration of Pb2+ ions by:
[H+]^2 = 2 * [Pb2+]
Now substitute back into the Q expression:
Q = 0.1403 / (2 * [Pb2+])
Next, we insert the values into the Nernst equation:
-0.1288V = 0.1260V - (8.314 J/(mol·K) * (25 + 273.15 K) / (2 * 96485 C/mol) * ln(Q))
Simplifying the equation, we can solve for ln(Q):
ln(Q) = -((0.1288 - 0.1260) / ((8.314 J/(mol·K) * (25 + 273.15 K) / (2 * 96485 C/mol))))
After obtaining the value for ln(Q), we can take the antilog to find Q:
Q = e^(ln(Q))
Now, we can substitute the value of Q back into the equation for [H+]^2:
[H+]^2 = 2 * [Pb2+]
Finally, we can solve for [H+], which is equal to the square root of (Q/2):
[H+] = √(Q/2)
Once [H+] is known, we can calculate the pH using the formula:
pH = -log10([H+])
Following these steps will allow us to determine the pH of the solution.