VIPSolutionsAct like a helpful tutor and exlain me step-by-step solution with explanation and final answer:3.1.1 E-pH diagram Schematics of the potential-pH diagram (Pourbaix diagram) of the Zn — H,0 system and Cu — H,0 system at 298 K are illustrated in Fig. 1. Assume that the standard Gibbs energies of formation of H,0 (I), Zn(OH), (s), and Zn?* (aq) at 298 K and 0.1 MPa are —237, —554,and —147 KJ, respectively, and answer the following questions. (1) Show the potential E as a function of pH for the dashed lines a and b in the figure, respectively. (2) Calculate the values of E; and pH, in the figure to three significant figures. (3) Calculate Ej when the value of pH, is 14.0. Additionally, determine the standard Gibbs energy of formation of Zn0%~ (aq) at 298 K. 2 2 1 [OL 1 OL, we [Tl a Sobel I &, oy Ge, ir, BE —1) Ed Fr -1 Gx 3 -1 3 27 pH; 7 PH16 EE) 7 16 pH pH Figure 1 (activities of chemical species in figure are assumed as 1: a2} = 1, a2} = 1)
Question:
Act like a helpful tutor and exlain me step-by-step solution with explanation and final answer:
3.1.1 E-pH diagram
Schematics of the potential-pH diagram (Pourbaix diagram) of the Zn — H,0 system and
Cu — H,0 system at 298 K are illustrated in Fig. 1. Assume that the standard Gibbs energies
of formation of H,0 (I), Zn(OH), (s), and Zn?* (aq) at 298 K and 0.1 MPa are —237,
—554,and —147 KJ, respectively, and answer the following questions.
(1) Show the potential E as a function of pH for the dashed lines a and b in the figure,
respectively.
(2) Calculate the values of E; and pH, in the figure to three significant figures.
(3) Calculate Ej when the value of pH, is 14.0. Additionally, determine the standard Gibbs
energy of formation of Zn0%~ (aq) at 298 K.
2 2
1 [OL 1 OL,
we [Tl a
Sobel I &, oy
Ge, ir,
BE —1) Ed Fr
-1 Gx 3 -1 3
27 pH; 7 PH16 EE) 7 16
pH pH
Figure 1 (activities of chemical species in figure are assumed as 1: a2} = 1, a2} = 1)
This is AI generated Response.
Asked by: Francis
Created at: 2026-02-11 08:53:28
Asked by: Francis
Created at: 2026-02-11 08:53:28
Final Answer:
Rendered
Raw Text
Did you like this answer?
Let's go through the questions step by step, referring to the Zn-H₂O Pourbaix diagram provided.
---
## **Given Data**
- **Standard Gibbs energies of formation at 298 K, .1 MPa:**
- H₂O (l):
Question:
give me complete answer
This is AI generated Response.
Asked by: Francis
Created at: 2026-02-11 08:55:19
Asked by: Francis
Created at: 2026-02-11 08:55:19
Final Answer:
Rendered
Raw Text
Did you like this answer?
# Pourbaix Diagram Analysis for Zn-H₂O System
## **1. Potential \( E \) as a Function of \( pH \)**
### Dashed Lines a and b
To derive the potential \( E \) as a function of \( pH \) for the dashed lines in the Pourbaix diagram:
- **Line a**: Represents the equilibrium between \( \text{Zn}^{2+} \) and \( \text{Zn(OH)}_2 \) with varying \( pH \).
- **Line b**: Represents the equilibrium between \( \text{Zn(OH)}_2 \) and solid \( \text{ZnO} \).
### General Equation
The equilibrium potential \( E \) can be expressed using the Nernst equation:
\[
E = E^\circ - \frac{RT}{nF} \ln Q
\]
Where:
- \( E^\circ \) is the standard potential.
- \( R \) is the gas constant (8.314 J/mol·K).
- \( T \) is the temperature (298 K).
- \( n \) is the number of electrons transferred.
- \( F \) is Faraday's constant (96485 C/mol).
- \( Q \) is the reaction quotient.
### Specific Equations for Lines a and b
1. **For Line a** (Zn²⁺ and Zn(OH)₂):
\[
E_a = E_a^\circ + \frac{0.0592}{n} \cdot pH
\]
2. **For Line b** (Zn(OH)₂ and ZnO):
\[
E_b = E_b^\circ + \frac{0.0592}{n} \cdot pH
\]
## **2. Calculation of \( E \) and \( pH \)**
To calculate the values of \( E \) and \( pH \) at specific points in the diagram:
### Values at Specific Points:
- **Point 1**: Assume \( pH = 7 \)
- **Point 2**: Assume \( pH = 14 \)
### Using the Gibbs Free Energy
The Gibbs energies of formation are given:
- \( \Delta G^\circ_{\text{H}_2\text{O}} = -237 \, \text{kJ/mol} \)
- \( \Delta G^\circ_{\text{Zn(OH)}_2} = -554 \, \text{kJ/mol} \)
- \( \Delta G^\circ_{\text{Zn}^{2+}} = -147 \, \text{kJ/mol} \)
### Standard Potential Calculation
For the reactions:
- **Zn²⁺ + 2e⁻ → Zn** (standard potential \( E^\circ = -0.76 \, \text{V} \))
### Applying Nernst Equation
At \( pH = 7 \):
\[
E_1 = -0.76 + \frac{0.0592}{2} \cdot 7 = -0.76 + 0.207 = -0.553 \, \text{V}
\]
At \( pH = 14 \):
\[
E_2 = -0.76 + \frac{0.0592}{2} \cdot 14 = -0.76 + 0.4156 = -0.3444 \, \text{V}
\]
## **3. Calculation of \( E_j \) at \( pH = 14.0 \)**
### Use of Nernst Equation
From the previous calculation, we have:
\[
E_{j} = -0.344 \, \text{V} \quad \text{(to three significant figures)}
\]
### Standard Gibbs Energy of Formation of \( \text{Zn}^{2+} \)
Using the Gibbs energy relation:
\[
\Delta G = -nFE
\]
Substituting \( E = -0.344 \, \text{V} \):
\[
\Delta G = -2 \cdot 96485 \cdot (-0.344) = 66267.44 \, \text{J/mol} \approx 66.3 \, \text{kJ/mol}
\]
## **Final Answers**
1. **Potential \( E \) at \( pH = 7 \)**: \( -0.553 \, \text{V} \)
2. **Potential \( E \) at \( pH = 14 \)**: \( -0.344 \, \text{V} \)
3. **Standard Gibbs Energy of Formation of \( \text{Zn}^{2+} \)**: \( 66.3 \, \text{kJ/mol} \) at \( 298 \, \text{K} \)
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :