Problem 97
Question
A magnesium battery can be constructed from an anode of magnesium metal and a cathode of molybdenum sulfide, \(\mathrm{Mo}_{3} \mathrm{S}_{4} .\) The standard reduction potentials of the electrode half-reactions are $$\mathrm{Mg}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Mg}(s) \quad E^{\circ}=-2.37 \mathrm{V}$$ \(\mathrm{Mg}^{2+}(a q)+\mathrm{Mo}_{3} \mathrm{S}_{4}(s)+2 \mathrm{e}^{-} \rightarrow \mathrm{MgMo}_{3} \mathrm{S}_{4}(s) \quad E^{\circ}=?\) a. If the standard cell potential for the battery is \(1.50 \mathrm{V}\) what is the value of \(E^{\circ}\) for the reduction of \(\mathrm{Mo}_{3} \mathrm{S}_{4} ?\) b. What are the apparent oxidation states of Mo in \(\mathrm{Mo}_{3} \mathrm{S}_{4}\) and in \(\mathrm{MgMo}_{3} \mathrm{S}_{4} ?\) "c. The electrolyte in the battery contains a complex magnesium salt, \(\mathrm{Mg}\left(\mathrm{AlCl}_{3} \mathrm{CH}_{3}\right)_{2} .\) Why is it necessary to include \(\mathrm{Mg}^{2+}\) ions in the electrolyte?
Step-by-Step Solution
Verified Answer
Answer: The standard reduction potential of Mo3S4 is 3.87 V. The oxidation state of Mo in Mo3S4 is 8/3, and in MgMo3S4, it is 2. Mg2+ ions are necessary in the electrolyte to complete the electrochemical reactions and maintain the electrical conductivity.
1Step 1: Using the Cell Potential formula to find the reduction potential of Mo3S4
According to the cell potential relationship, the difference between the reduction potential values is equal to the cell potential (E°cell = E°cathode - E°anode). Let's find the unknown reduction potential value (EMo3S4) using the given cell potential.
E°cell = 1.50 V (given)
E°Mg = -2.37 V (given)
Using the relationship:
1.50 = E°Mo3S4 - (-2.37)
2Step 2: Calculate the value of E°Mo3S4
Solving for E°Mo3S4:
E°Mo3S4 = 1.50 + 2.37
E°Mo3S4 = 3.87 V
The standard reduction potential for Mo3S4 is 3.87 V.
3Step 3: Finding oxidation state of Mo in Mo3S4 and MgMo3S4
Let's assign the oxidation state of Mo in Mo3S4 as x. As we know sulfur has an oxidation state of -2, we can set up the following equation:
3x + (4 * (-2)) = 0 (The whole compound should have a total charge of zero)
3x - 8 = 0
Solve for x:
3x = 8
x = 8/3
In Mo3S4, the oxidation state of Mo = 8/3
Let's do the same for MgMo3S4:
Mg has an oxidation state of +2. Calculate the oxidation state of Mo in MgMo3S4:
3y + (4 * (-2)) + 2 = 0 (The whole compound should have a total charge of zero)
3y - 6 = 0
Solve for y:
y = 2
In MgMo3S4, the oxidation state of Mo = 2
4Step 4: Explaining the necessity of Mg2+ ions in the electrolyte
The electrolyte solution contains Mg2+ ions because magnesium is being oxidized in the anode. During the electrochemical process in the battery, Mg is being converted to Mg2+ ions, releasing 2 electrons in the process. These ions should be available in the electrolyte to complete the chemical reactions and, thus maintain the electrical conductivity.
In conclusion:
a. The standard reduction potential of Mo3S4 is 3.87 V
b. The oxidation state of Mo in Mo3S4 is 8/3, and in MgMo3S4 it is 2
c. Mg2+ ions are necessary in the electrolyte to complete the electrochemical reactions and maintain the electrical conductivity.
Key Concepts
Standard Reduction PotentialOxidation StatesElectrochemical Cell
Standard Reduction Potential
The standard reduction potential is a measure of the tendency of a chemical species to gain electrons and be reduced. When we discuss electrochemical cells, we often need to determine the overall cell potential, which is the voltage or electrical potential difference between two electrodes.
To find this value, we use the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
In our example with the magnesium battery, we know:
\[ 1.50 = E^{\circ}_{\text{Mo}_{3}\text{S}_{4}} - (-2.37) \]
Solving this gives us \( E^{\circ}_{\text{Mo}_{3}\text{S}_{4}} = 3.87 \) V. This information tells us how efficiently the \( \text{Mo}_{3}\text{S}_{4} \) gains electrons in this setup.
To find this value, we use the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
In our example with the magnesium battery, we know:
- The standard cell potential \( E^{\circ}_{\text{cell}} \) is 1.50 V.
- The standard reduction potential for the magnesium anode \( E^{\circ}_{\text{Mg}} \) is -2.37 V.
\[ 1.50 = E^{\circ}_{\text{Mo}_{3}\text{S}_{4}} - (-2.37) \]
Solving this gives us \( E^{\circ}_{\text{Mo}_{3}\text{S}_{4}} = 3.87 \) V. This information tells us how efficiently the \( \text{Mo}_{3}\text{S}_{4} \) gains electrons in this setup.
Oxidation States
Oxidation states, also known as oxidation numbers, are a way to keep track of electrons during chemical reactions. They help us understand which elements lose or gain electrons, often providing insights into redox reactions.
For the compound \( \text{Mo}_{3}\text{S}_{4} \), we calculate the oxidation state of molybdenum (Mo) as follows:
Now, looking at \( \text{MgMo}_{3}\text{S}_{4} \), where magnesium (Mg) has an oxidation state of +2, we adjust the equation:
\[ 3y + (4 \times -2) + 2 = 0 \]
Solving for \( y \) gives 2, meaning the oxidation state of Mo here is 2. This change suggests that Mo undergoes a reduction process when forming the compound.
For the compound \( \text{Mo}_{3}\text{S}_{4} \), we calculate the oxidation state of molybdenum (Mo) as follows:
- Assume the oxidation state of \( \text{Mo} \) is \( x \).
- Since sulfur (S) has a known oxidation state of -2, the equation becomes:
\[ 3x + (4 \times -2) = 0 \]
Now, looking at \( \text{MgMo}_{3}\text{S}_{4} \), where magnesium (Mg) has an oxidation state of +2, we adjust the equation:
\[ 3y + (4 \times -2) + 2 = 0 \]
Solving for \( y \) gives 2, meaning the oxidation state of Mo here is 2. This change suggests that Mo undergoes a reduction process when forming the compound.
Electrochemical Cell
An electrochemical cell is a device that can generate electrical energy from chemical reactions or use electrical energy to drive chemical reactions. In the case of a magnesium battery, we leverage these reactions in a controlled manner.
The magnesium acts as the anode, undergoing oxidation. This process involves losing electrons and forming \( \text{Mg}^{2+} \) ions, which are crucial for the ongoing reaction. On the other hand, \( \text{Mo}_{3}\text{S}_{4} \) serves as the cathode, where the reduction occurs by gaining electrons.
To ensure smooth operation:
The magnesium acts as the anode, undergoing oxidation. This process involves losing electrons and forming \( \text{Mg}^{2+} \) ions, which are crucial for the ongoing reaction. On the other hand, \( \text{Mo}_{3}\text{S}_{4} \) serves as the cathode, where the reduction occurs by gaining electrons.
To ensure smooth operation:
- The electrolyte must contain \( \text{Mg}^{2+} \) ions. This helps balance the charges and allows electrons to flow between the anode and cathode efficiently.
- Connectivity maintained by the electrolyte ensures the flow of ions and prevents a build-up of charges that would stop the cell from functioning.
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