Answer :
Let's approach this problem step-by-step.
### Step 1: Understand the Reaction
We are given a galvanic cell where tin metal (Sn) is placed in a solution containing [tex]\( \text{Re}^{+} \)[/tex] ions at unit molarity. The half-reactions involved and their standard electrode potentials are:
[tex]\[ \begin{array}{l} E_{\text{Sn}^{2+}/\text{Sn}}^0=-0.44 \, \text{V} \\ E_{\text{Re}^{+}/\text{Re}}^0=0.77 \, \text{V} \\ \end{array} \][/tex]
### Step 2: Identify Oxidation and Reduction
For a galvanic cell, the species with the lower (or more negative) standard electrode potential will be oxidized (lose electrons), and the species with the higher (or more positive) standard electrode potential will be reduced (gain electrons).
From the given data:
- The standard electrode potential for [tex]\(\text{Sn}^{2+}/\text{Sn}\)[/tex] is [tex]\(-0.44 \, \text{V}\)[/tex], which is more negative.
- The standard electrode potential for [tex]\(\text{Re}^{+}/\text{Re}\)[/tex] is [tex]\(0.77 \, \text{V}\)[/tex], which is more positive.
Thus:
- Tin (Sn) will be oxidized (lose electrons):
[tex]\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \][/tex]
- Rhenium ion ([tex]\(\text{Re}^{+}\)[/tex]) will be reduced (gain electrons):
[tex]\[ \text{Re}^{+} + e^- \rightarrow \text{Re} \][/tex]
### Step 3: Calculate the Standard Cell Potential ([tex]\(E_{\text{cell}}^{\circ}\)[/tex])
The standard cell potential is calculated using the standard electrode potentials of the reduction and oxidation half-reactions forms the cell potential equation:
[tex]\[ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \][/tex]
Given:
- Standard electrode potential for the cathode (reduction, [tex]\(\text{Re}^{+}/\text{Re}\)[/tex]): [tex]\(E_{\text{Re}^{+}/\text{Re}}^{\circ} = 0.77 \, \text{V}\)[/tex]
- Standard electrode potential for the anode (oxidation, [tex]\(\text{Sn}^{2+}/\text{Sn}\)[/tex]): [tex]\(E_{\text{Sn}^{2+}/\text{Sn}}^{\circ} = -0.44 \, \text{V}\)[/tex]
Thus:
[tex]\[ E_{\text{cell}}^{\circ} = 0.77 \, \text{V} - (-0.44 \, \text{V}) = 0.77 \, \text{V} + 0.44 \, \text{V} = 1.21 \, \text{V} \][/tex]
### Step 4: Write the Cell Notation (Cell Diagram)
The cell notation for a galvanic cell is written to represent the flow from the anode to the cathode, including the solutions involved.
For the reaction:
- Oxidation at the anode (left side): [tex]\(\text{Sn} \rightarrow \text{Sn}^{2+}\)[/tex]
- Reduction at the cathode (right side): [tex]\(\text{Re}^{+} \rightarrow \text{Re}\)[/tex]
The cell notation is:
[tex]\[ \text{Sn} \, | \, \text{Sn}^{2+} \, || \, \text{Re}^{+} \, | \, \text{Re} \][/tex]
### Summary of the Solution
- Standard Electrode Potentials:
[tex]\[ E_{\text{Sn}^{2+}/\text{Sn}}^0 = -0.44 \, \text{V} \][/tex]
[tex]\[ E_{\text{Re}^{+}/\text{Re}}^0 = 0.77 \, \text{V} \][/tex]
- Standard Cell Potential ([tex]\(E_{\text{cell}}^{\circ}\)[/tex]):
[tex]\[ E_{\text{cell}}^{\circ} = 1.21 \, \text{V} \][/tex]
- Cell Notation:
[tex]\[ \text{Sn} \, | \, \text{Sn}^{2+} \, || \, \text{Re}^{+} \, | \, \text{Re} \][/tex]
### Step 1: Understand the Reaction
We are given a galvanic cell where tin metal (Sn) is placed in a solution containing [tex]\( \text{Re}^{+} \)[/tex] ions at unit molarity. The half-reactions involved and their standard electrode potentials are:
[tex]\[ \begin{array}{l} E_{\text{Sn}^{2+}/\text{Sn}}^0=-0.44 \, \text{V} \\ E_{\text{Re}^{+}/\text{Re}}^0=0.77 \, \text{V} \\ \end{array} \][/tex]
### Step 2: Identify Oxidation and Reduction
For a galvanic cell, the species with the lower (or more negative) standard electrode potential will be oxidized (lose electrons), and the species with the higher (or more positive) standard electrode potential will be reduced (gain electrons).
From the given data:
- The standard electrode potential for [tex]\(\text{Sn}^{2+}/\text{Sn}\)[/tex] is [tex]\(-0.44 \, \text{V}\)[/tex], which is more negative.
- The standard electrode potential for [tex]\(\text{Re}^{+}/\text{Re}\)[/tex] is [tex]\(0.77 \, \text{V}\)[/tex], which is more positive.
Thus:
- Tin (Sn) will be oxidized (lose electrons):
[tex]\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \][/tex]
- Rhenium ion ([tex]\(\text{Re}^{+}\)[/tex]) will be reduced (gain electrons):
[tex]\[ \text{Re}^{+} + e^- \rightarrow \text{Re} \][/tex]
### Step 3: Calculate the Standard Cell Potential ([tex]\(E_{\text{cell}}^{\circ}\)[/tex])
The standard cell potential is calculated using the standard electrode potentials of the reduction and oxidation half-reactions forms the cell potential equation:
[tex]\[ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \][/tex]
Given:
- Standard electrode potential for the cathode (reduction, [tex]\(\text{Re}^{+}/\text{Re}\)[/tex]): [tex]\(E_{\text{Re}^{+}/\text{Re}}^{\circ} = 0.77 \, \text{V}\)[/tex]
- Standard electrode potential for the anode (oxidation, [tex]\(\text{Sn}^{2+}/\text{Sn}\)[/tex]): [tex]\(E_{\text{Sn}^{2+}/\text{Sn}}^{\circ} = -0.44 \, \text{V}\)[/tex]
Thus:
[tex]\[ E_{\text{cell}}^{\circ} = 0.77 \, \text{V} - (-0.44 \, \text{V}) = 0.77 \, \text{V} + 0.44 \, \text{V} = 1.21 \, \text{V} \][/tex]
### Step 4: Write the Cell Notation (Cell Diagram)
The cell notation for a galvanic cell is written to represent the flow from the anode to the cathode, including the solutions involved.
For the reaction:
- Oxidation at the anode (left side): [tex]\(\text{Sn} \rightarrow \text{Sn}^{2+}\)[/tex]
- Reduction at the cathode (right side): [tex]\(\text{Re}^{+} \rightarrow \text{Re}\)[/tex]
The cell notation is:
[tex]\[ \text{Sn} \, | \, \text{Sn}^{2+} \, || \, \text{Re}^{+} \, | \, \text{Re} \][/tex]
### Summary of the Solution
- Standard Electrode Potentials:
[tex]\[ E_{\text{Sn}^{2+}/\text{Sn}}^0 = -0.44 \, \text{V} \][/tex]
[tex]\[ E_{\text{Re}^{+}/\text{Re}}^0 = 0.77 \, \text{V} \][/tex]
- Standard Cell Potential ([tex]\(E_{\text{cell}}^{\circ}\)[/tex]):
[tex]\[ E_{\text{cell}}^{\circ} = 1.21 \, \text{V} \][/tex]
- Cell Notation:
[tex]\[ \text{Sn} \, | \, \text{Sn}^{2+} \, || \, \text{Re}^{+} \, | \, \text{Re} \][/tex]
