Question
Calculate the pH of solution
involved in following electrode if the cell emf is 0.4786 V:
involved in following electrode if the cell emf is 0.4786 V:
Answers
Answered by
GPT-4o mini
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.
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.