The following acidic galvanic cell is set-up with the following overall cell reaction 14 H* (aq) + Cr207 (aq) 6 Ag(s) → 6 Ag"(aq) + 2 Cr³* (aq) +7 H2O0) V Porous Barrier Pt Ag 0.22 M Na2Cr207 0.20 M AGNO3 0.25 M Cr(NO3)3 If the cell voltage is 0.257 V, what is the pH of the cell?
The following acidic galvanic cell is set-up with the following overall cell reaction 14 H* (aq) + Cr207 (aq) 6 Ag(s) → 6 Ag"(aq) + 2 Cr³* (aq) +7 H2O0) V Porous Barrier Pt Ag 0.22 M Na2Cr207 0.20 M AGNO3 0.25 M Cr(NO3)3 If the cell voltage is 0.257 V, what is the pH of the cell?
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![### Acidic Galvanic Cell Reaction and pH Calculation
#### Galvanic Cell Setup and Reaction
The following acidic galvanic cell is set up with the overall cell reaction below:
\[ 14 \text{H}^+ _{(aq)} + \text{Cr}_2\text{O}_7^{2-} _{(aq)} + 6 \text{Ag}_{(s)} \rightarrow 6 \text{Ag}^+ _{(aq)} + 2 \text{Cr}^{3+} _{(aq)} + 7 \text{H}_2\text{O}_{(l)} \]
#### Diagram Explanation
The diagram depicts a galvanic cell with two half-cells separated by a porous barrier:
- **Left Half-Cell:** Contains a platinum (Pt) electrode immersed in a solution with concentrations of 0.22 M Na\(_2\)Cr\(_2\)O\(_7\) and 0.25 M Cr(NO\(_3\))\(_3\).
- **Right Half-Cell:** Contains a silver (Ag) electrode in a 0.20 M AgNO\(_3\) solution.
- **Porous Barrier:** Allows ion exchange between the two half-cells without mixing the solutions.
- **Voltage Measurement:** A voltmeter is connected to both electrodes to measure the potential difference (cell voltage).
#### Problem Statement
If the cell voltage is 0.257 V, what is the pH of the cell?
By following the principles of electrochemistry and using the provided data, one can calculate the pH of the cell.
This setup illustrates how a galvanic cell operates, with the specified reaction and given conditions, facilitating the understanding of redox reactions, electrode potentials, and pH calculations in an educational context.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F43ef5448-495e-49e8-9cec-405cabc5c22f%2Fb18f0eb4-89ae-4d94-96d6-f13721372a0a%2F3lya2yd.jpeg&w=3840&q=75)
Transcribed Image Text:### Acidic Galvanic Cell Reaction and pH Calculation
#### Galvanic Cell Setup and Reaction
The following acidic galvanic cell is set up with the overall cell reaction below:
\[ 14 \text{H}^+ _{(aq)} + \text{Cr}_2\text{O}_7^{2-} _{(aq)} + 6 \text{Ag}_{(s)} \rightarrow 6 \text{Ag}^+ _{(aq)} + 2 \text{Cr}^{3+} _{(aq)} + 7 \text{H}_2\text{O}_{(l)} \]
#### Diagram Explanation
The diagram depicts a galvanic cell with two half-cells separated by a porous barrier:
- **Left Half-Cell:** Contains a platinum (Pt) electrode immersed in a solution with concentrations of 0.22 M Na\(_2\)Cr\(_2\)O\(_7\) and 0.25 M Cr(NO\(_3\))\(_3\).
- **Right Half-Cell:** Contains a silver (Ag) electrode in a 0.20 M AgNO\(_3\) solution.
- **Porous Barrier:** Allows ion exchange between the two half-cells without mixing the solutions.
- **Voltage Measurement:** A voltmeter is connected to both electrodes to measure the potential difference (cell voltage).
#### Problem Statement
If the cell voltage is 0.257 V, what is the pH of the cell?
By following the principles of electrochemistry and using the provided data, one can calculate the pH of the cell.
This setup illustrates how a galvanic cell operates, with the specified reaction and given conditions, facilitating the understanding of redox reactions, electrode potentials, and pH calculations in an educational context.
Expert Solution
Step 1
The Nernst equation is shown below:
Where;
Ecell = electrode potential at any condition
E0cell = standard electrode potential
R = gas constant
T = temperature
n = no. of electrons transferred
F = Faraday's constant
Q = reaction quotient
Step 2
The given cell reaction is shown below:
Step 3
The reduction half cell reaction is shown below:
The oxidation half cell reaction is shown below:
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