Calculate the pH of solution

involved in following electrode if the cell emf is 0.4786 V:

1 answer

To calculate the pH of a solution from the cell electromotive force (emf), we typically use the Nernst equation. However, we need more information about the specific half-reactions taking place in the electrochemical cell to accurately calculate the pH.

If there's a standard hydrogen electrode (SHE) involved, the half-reaction for the reduction of hydrogen ions can be given as: \[ \text{2H}^+ + 2e^- \leftrightarrow \text{H}_2(g) \] At standard conditions (1 M H\(^+\) concentration), the standard electrode potential (\(E^\circ\)) for this half-reaction is 0 V.

The Nernst equation for this half-reaction is: \[ E = E^\circ - \frac{RT}{nF} \ln Q \] where:

  • \(E\) is the cell potential (0.4786 V)
  • \(E^\circ\) is the standard electrode potential (0 V for SHE)
  • \(R\) is the universal gas constant (8.314 J/(mol·K))
  • \(T\) is the temperature in Kelvin (assumed to be 298 K unless specified otherwise)
  • \(n\) is the number of moles of electrons transferred in the half-reaction (2 for the reduction of hydrogen ions)
  • \(F\) is Faraday's constant (96485 C/mol)
  • \(Q\) is the reaction quotient, which for the hydrogen half-reaction can be expressed as \(Q = 1 / [\text{H}^+]^2\).

Rearranging the Nernst equation gets us: \[ 0.4786 = 0 - \frac{(8.314)(298)}{(2)(96485)} \ln \left(\frac{1}{[\text{H}^+]^2}\right) \]

Calculating the constants: \[ \frac{(8.314)(298)}{(2)(96485)} = \frac{2478.172}{192970} \approx 0.01284 \]

Now plug this back into our equation: \[ 0.4786 = -0.01284 \ln \left(\frac{1}{[\text{H}^+]^2}\right) \]

To simplify: \[ 0.4786 = 0.01284 \ln \left([\text{H}^+]^2\right) \]

This means we can express this as: \[ \ln \left([\text{H}^+]^2\right) = \frac{0.4786}{0.01284} \] \[ \ln \left([\text{H}^+]^2\right) \approx 37.27 \]

Now we exponentiate both sides: \[ [\text{H}^+]^2 \approx e^{37.27} \]

Calculate \(e^{37.27}\): \[ [\text{H}^+]^2 \approx 1.557 \times 10^{16} \] \[ [\text{H}^+] \approx \sqrt{1.557 \times 10^{16}} \approx 3.95 \times 10^8 \]

Finally, the pH is calculated as: \[ \text{pH} = -\log([\text{H}^+]) \approx -\log(3.95 \times 10^8) \approx -8.597 \]

Since we usually express pH within a standard range, you should consider revising some experimental or initial parameters as this suggests a very acidic solution beyond typical levels. In real scenarios, pH usually ranges from 0 to 14.

Thus, please provide the specific half-reactions or any additional context if available to correct this calculation or to ensure proper factors are taken into consideration.