Please answer the attached question: ---### Standard Reduction PotentialsThe table below provides standard reduction potentials (E°) for various redox couples. The standard reduction potential is an important concept in electrochemistry, indicating the tendency of a chemical species to gain electrons and thereby be reduced. The more positive the reduction potential, the greater the species' affinity for electrons, and thus its tendency to be reduced is higher.#### Table: Standard Reduction Potentials| **Couple** | | **E° (Volts)** | **Couple** | | **E° (Volts)** ||------------------|-------------------|-------------|----------------|---------------|-------------|| F₂ | ---- HF (H⁺) | +3.03 | SO₄²⁻ | ---- H₂SO₃ (H⁺) | +0.20 || F₂ | ---- F⁻ | +2.87 | Sn⁴⁺ | ---- Sn²⁺ | +0.15 || S₂O₈²⁻ | ---- SO₄²⁻ | +2.05 | S | ---- H₂S (H⁺) | +0.141 || BiO₃⁻ | ---- Bi³⁺ | +2.0 | Hg₂Br₂ | ---- Hg (Br⁻) | +0.140 || H₂O₂ | ---- H₂O (H⁺) | +1.78 | AgBr | ---- Ag (Br⁻) | +0.0713 || PbO₂ | PbSO₄ (H⁺, SO₄²⁻) | +1.685 | H⁺ | ---- H₂ | +0.0000 || Ce⁴⁺ | ---- Ce³⁺ | +1.61 | Pb²⁺ | ---- Pb | -0.126 || MnO₄⁻ | ---- Mn²⁺ (H⁺) | +1.491 | Sn²⁺ | ---- Sn | -0.136 || ClO₃⁻ | ---- Cl⁻ | +1.47 | AgI ### Acidic Galvanic Cell Setup and pH Determination#### The Cell ReactionThe acidic galvanic cell is set up with the following overall cell reaction:\[14 \text{H}^+_{\text{(aq)}} + \text{Cr}_2\text{O}_7^{2-} _{\text{(aq)}}+ 6 \text{Ag}_{\text{(s)}} \rightarrow 6 \text{Ag}^+_{\text{(aq)}} + 2 \text{Cr}^{3+}_{\text{(aq)}} + 7 \text{H}_2\text{O}_{\text{(l)}}\]#### Diagram DescriptionThe diagram represents a galvanic cell with:- A porous barrier separating the two half-cells.- The left half-cell contains a Platinum (Pt) electrode immersed in a solution of 0.22 M Na\(_2\)Cr\(_2\)O\(_7\) and 0.25 M Cr(NO\(_3\))\(_3\).- The right half-cell contains a Silver (Ag) electrode immersed in a 0.20 M AgNO\(_3\) solution.- The setup includes a voltmeter (V) connected between the two electrodes, indicating the cell voltage.#### Problem Statement"If the cell voltage is 0.257 V, what is the pH of the cell?"To solve this, you would use the Nernst equation and other relevant electrochemical principles to find the pH of the solution in the galvanic cell based on the given conditions.

Answered: The following acidic galvanic cell is…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

Please answer the attached question:

---

### Standard Reduction Potentials

The table below provides standard reduction potentials (E°) for various redox couples. The standard reduction potential is an important concept in electrochemistry, indicating the tendency of a chemical species to gain electrons and thereby be reduced. The more positive the reduction potential, the greater the species' affinity for electrons, and thus its tendency to be reduced is higher.

#### Table: Standard Reduction Potentials

| **Couple** | | **E° (Volts)** | **Couple** | | **E° (Volts)** |
|------------------|-------------------|-------------|----------------|---------------|-------------|
| F₂ | ---- HF (H⁺) | +3.03 | SO₄²⁻ | ---- H₂SO₃ (H⁺) | +0.20 |
| F₂ | ---- F⁻ | +2.87 | Sn⁴⁺ | ---- Sn²⁺ | +0.15 |
| S₂O₈²⁻ | ---- SO₄²⁻ | +2.05 | S | ---- H₂S (H⁺) | +0.141 |
| BiO₃⁻ | ---- Bi³⁺ | +2.0 | Hg₂Br₂ | ---- Hg (Br⁻) | +0.140 |
| H₂O₂ | ---- H₂O (H⁺) | +1.78 | AgBr | ---- Ag (Br⁻) | +0.0713 |
| PbO₂ | PbSO₄ (H⁺, SO₄²⁻) | +1.685 | H⁺ | ---- H₂ | +0.0000 |
| Ce⁴⁺ | ---- Ce³⁺ | +1.61 | Pb²⁺ | ---- Pb | -0.126 |
| MnO₄⁻ | ---- Mn²⁺ (H⁺) | +1.491 | Sn²⁺ | ---- Sn | -0.136 |
| ClO₃⁻ | ---- Cl⁻ | +1.47 | AgI

### Acidic Galvanic Cell Setup and pH Determination

#### The Cell Reaction
The acidic galvanic cell is set up with the following overall cell reaction:

\[14 \text{H}^+_{\text{(aq)}} + \text{Cr}_2\text{O}_7^{2-} _{\text{(aq)}}+ 6 \text{Ag}_{\text{(s)}} \rightarrow 6 \text{Ag}^+_{\text{(aq)}} + 2 \text{Cr}^{3+}_{\text{(aq)}} + 7 \text{H}_2\text{O}_{\text{(l)}}\]

#### Diagram Description
The diagram represents a galvanic cell with:

- A porous barrier separating the two half-cells.
- The left half-cell contains a Platinum (Pt) electrode immersed in a solution of 0.22 M Na\(_2\)Cr\(_2\)O\(_7\) and 0.25 M Cr(NO\(_3\))\(_3\).
- The right half-cell contains a Silver (Ag) electrode immersed in a 0.20 M AgNO\(_3\) solution.
- The setup includes a voltmeter (V) connected between the two electrodes, indicating the cell voltage.

#### Problem Statement
"If the cell voltage is 0.257 V, what is the pH of the cell?"

To solve this, you would use the Nernst equation and other relevant electrochemical principles to find the pH of the solution in the galvanic cell based on the given conditions.

Expert Solution

Introduction:

The potential across any specified cell is computed using all species' concentration and its standard potential. This expression is also recognized as "Nernst equation" and it also contains the term denoting electrons transferred.

Given:

The temperature is 25 degree Celsius.

The concentration of silver ions is 0.20 M.

The concentration of chromium ions is 0.25 M.

The concentration of chromate ions is 0.22 M.

The cell voltage is 0.257 V.

Solution:

The given reaction is as follows:

The oxidation half-reaction at anode is shown below.

The reduction half-reaction at cathode is shown below.

Multiply oxidation-half with 6 and then add oxidation and reduction half-reactions to obtain net reaction as shown below.

The formula to calculate pH is as follows:

The concentration of hydronium ions is calculated Nernst equation as shown below.

Here,

The number of electrons transfer in the reaction is “n”.

The cell potential is “E_cell”.

The standard cell potential is “”.

It clear from the half-reactions that the number of electrons transferred is 6 (n=6).

The formula to calculate standard cell potential is shown below.

Here,

The electrode potential of cathode is +1.33 V.

The electrode potential of anode is +0.7996 V.

Substitute the values in equation (III).

Suppose the concentration of hydrogen ions is x.

Substitute all known values in equation (II).

On further solving the equation,

Solving the equation for the value of x:

Substitute the concentration of hydrogen ions in equation (I).

Step by step

Solved in 3 steps with 12 images

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions