To solve the problem, we need to analyze the electrolysis of water occurring in the voltaic cell and how it affects the pH of the buffer solutions in each compartment.
### Step-by-Step Solution:
1. **Identify the Reactions at the Electrodes**:
- At the **anode** (oxidation reaction), water is oxidized to produce oxygen gas, protons (H⁺), and electrons:
\[
2H_2O \rightarrow O_2 + 4H^+ + 4e^-
\]
- At the **cathode** (reduction reaction), water is reduced to produce hydrogen gas and hydroxide ions (OH⁻):
\[
2H_2O + 2e^- \rightarrow H_2 + 2OH^-
\]
2. **Calculate the Total Charge Passed**:
- The current (I) is given as 1.5 A and the time (t) is 20 minutes (which is 1200 seconds).
- Total charge (Q) can be calculated using the formula:
\[
Q = I \times t = 1.5 \, \text{A} \times 1200 \, \text{s} = 1800 \, \text{C}
\]
3. **Determine Moles of Electrons Transferred**:
- Using Faraday's constant (F = 96500 C/mol), we can find the number of moles of electrons (n):
\[
n = \frac{Q}{F} = \frac{1800 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.01865 \, \text{mol}
\]
4. **Calculate the Change in H⁺ and OH⁻ Concentrations**:
- At the anode, for every 4 moles of electrons, 4 moles of H⁺ are produced. Therefore, the moles of H⁺ produced:
\[
\text{Moles of H}^+ = n = 0.01865 \, \text{mol}
\]
- At the cathode, for every 2 moles of electrons, 2 moles of OH⁻ are produced. Therefore, the moles of OH⁻ produced:
\[
\text{Moles of OH}^- = n = 0.01865 \, \text{mol}
\]
5. **Calculate the Change in pH**:
- The volume of the buffer solution in each compartment is 100 mL (0.1 L).
- The concentration of H⁺ produced at the anode:
\[
[H^+] = \frac{0.01865 \, \text{mol}}{0.1 \, \text{L}} = 0.1865 \, \text{M}
\]
- The concentration of OH⁻ produced at the cathode:
\[
[OH^-] = \frac{0.01865 \, \text{mol}}{0.1 \, \text{L}} = 0.1865 \, \text{M}
\]
6. **Determine the pH Changes**:
- The pH at the anode will decrease due to the increase in H⁺ concentration:
\[
\text{pH}_{\text{anode}} = -\log(0.1865) \approx 0.728
\]
- The pH at the cathode will increase due to the increase in OH⁻ concentration:
\[
\text{pOH}_{\text{cathode}} = -\log(0.1865) \approx 0.728 \quad \Rightarrow \quad \text{pH}_{\text{cathode}} = 14 - 0.728 \approx 13.272
\]
7. **Final pH of the Buffer**:
- Since the buffer consists of NH₃ and NH₄⁺, the overall pH will be influenced by the changes in both compartments. However, the average pH can be approximated as:
\[
\text{Average pH} = \frac{\text{pH}_{\text{anode}} + \text{pH}_{\text{cathode}}}{2} = \frac{0.728 + 13.272}{2} \approx 7.0
\]
### Conclusion:
The pH of the buffer after electrolysis will be approximately 7.0.
