Question

In nature, many electron transfer reactions take place which are catalyzed by enzymes. Two such half-cell reactions are:

$2H^+(aq) + 2e^- \longrightarrow H_2 \text{ (g, 1 atm)}$

$O_2 \text{ (g, 1 atm)} + 4H^+(aq) + 4e^- \longrightarrow 2H_2O \text{ (l)} \quad E^o = 1.23 \text{ V}$

The reduction potentials for the two half-cell reactions respectively at the physiological pH 7 are:

• A. 0.0 V and +0.82 V
• B. -0.41 V and +0.82 V
• C. 0.0 V and +1.23 V
• D. -0.41 V and +1.23 V

Answer 1

Answer: (B)

The reduction potentials for the two half-cell reactions at physiological pH 7 are approximately -0.41 V and +0.82 V, respectively.

Answer 2

To find the reduction potentials at physiological pH 7, we use the Nernst equation:

$E = E^o - \frac{0.0591}{n} \log Q$

where $E^o$ is the standard reduction potential, $n$ is the number of electrons transferred, and $Q$ is the reaction quotient. At pH 7, the hydrogen ion concentration $[H^+]$ is $10^{-7} \text{ M}$.

For the first half-cell reaction:

$2H^+(aq) + 2e^- \longrightarrow H_2 \text{ (g, 1 atm)}$

1. Standard Reduction Potential ($E^o$): By definition, the standard reduction potential for the Standard Hydrogen Electrode (SHE) is $E^o = 0.0 \text{ V}$.

2. Number of electrons ($n$): From the balanced reaction, $n = 2$.

3. Reaction Quotient ($Q$): $Q = \frac{P_{H_2}}{[H^+]^2}$

   Given $P_{H_2} = 1 \text{ atm}$ and $[H^+] = 10^{-7} \text{ M}$.

   $Q = \frac{1}{(10^{-7})^2} = \frac{1}{10^{-14}} = 10^{14}$

4. Calculate $E$ at pH 7:

   $E_1 = E^o - \frac{0.0591}{n} \log Q$

   $E_1 = 0.0 - \frac{0.0591}{2} \log (10^{14})$

   $E_1 = 0.0 - \frac{0.0591}{2} \times 14$

   $E_1 = 0.0 - 0.0591 \times 7$

   $E_1 = -0.4137 \text{ V} \approx -0.41 \text{ V}$

For the second half-cell reaction:

$O_2 \text{ (g, 1 atm)} + 4H^+(aq) + 4e^- \longrightarrow 2H_2O \text{ (l)}$

1. Standard Reduction Potential ($E^o$): Given $E^o = 1.23 \text{ V}$.

2. Number of electrons ($n$): From the balanced reaction, $n = 4$.

3. Reaction Quotient ($Q$): $Q = \frac{1}{P_{O_2}[H^+]^4}$ (activity of pure liquid water is 1)

   Given $P_{O_2} = 1 \text{ atm}$ and $[H^+] = 10^{-7} \text{ M}$.

   $Q = \frac{1}{1 \times (10^{-7})^4} = \frac{1}{10^{-28}} = 10^{28}$

4. Calculate $E$ at pH 7:

   $E_2 = E^o - \frac{0.0591}{n} \log Q$

   $E_2 = 1.23 - \frac{0.0591}{4} \log (10^{28})$

   $E_2 = 1.23 - \frac{0.0591}{4} \times 28$

   $E_2 = 1.23 - 0.0591 \times 7$

   $E_2 = 1.23 - 0.4137 \text{ V}$

   $E_2 = 0.8163 \text{ V} \approx +0.82 \text{ V}$

Thus, the reduction potentials for the two half-cell reactions at physiological pH 7 are approximately -0.41 V and +0.82 V, respectively.