Problem 19 - Chemistry: Principles and Reactions

Question

Calculate \(E^{\circ}\) for the following voltaic cells: (a) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}(s)\) (b) \(\mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q)+\mathrm{S}(s) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{S}^{2-}(a q)\) (c) an \(\mathrm{Ag}-\mathrm{Ag}^{+}\) half-cell and an \(\mathrm{Au}-\mathrm{AuCl}_{4}^{-}\) half-cell.

Step-by-Step Solution

Answer

Question: Calculate the standard cell potentials for the following voltaic cells: (a) MnO₂(s) + 4H⁺(aq) + 2I⁻(aq) → Mn²⁺(aq) + 2H₂O(l) + I₂(s) (b) S(s) + 2H₂(g) → S²⁻(aq) + 4H⁺(aq) (c) half-cell 1: Ag⁺ → Ag half-cell 2: Au³⁺ + 4Cl⁻ → AuCl₄⁻ Answer: The standard cell potentials for the given voltaic cells are as follows: (a) \(E^{\circ}_a = 0.69\,\text{V}\) (b) \(E^{\circ}_b = -0.14\,\text{V}\) (c) \(E^{\circ}_c = 0.13\,\text{V}\)

Step 1: Identify the half-reactions for (a) and (b)

First, we need to split the given reactions (a) and (b) into their respective half-reactions. For (a), we can see that MnO2 is reduced, and I⁻ is oxidized. For (b), H₂ is oxidized and S is reduced. (a) Reduction half-reaction: MnO₂(s) + 4H⁺(aq) + 2e⁻ → Mn²⁺(aq) + 2H₂O(l) Oxidation half-reaction: 2I⁻(aq) → I₂(s) + 2e⁻ (b) Reduction half-reaction: S(s) + 2e⁻ → S²⁻(aq) Oxidation half-reaction: H₂(g) → 2H⁺(aq) + 2e⁻ Now that we have the half-reactions for (a) and (b), we will look up their standard reduction potentials.

Step 2: Determine the standard reduction potentials

Using a standard reduction potential table, we can find the following values for the half-reactions: For (a) Reduction: \(E_{red,a}^\circ(MnO_2) = 1.23\,\text{V}\) For (a) Oxidation: \(E_{red,a}^\circ(I^-) = 0.54\,\text{V}\) For (b) Reduction: \(E_{red,b}^\circ(S) = -0.14\,\text{V}\) For (b) Oxidation: \(E_{red,b}^\circ(H_2) = 0\,\text{V}\)

Step 3: Calculate the standard cell potentials

To calculate the standard cell potentials, we can now use the following formula: \(E^{\circ} = E_{red}^\circ(\text{reduction}) - E_{red}^\circ(\text{oxidation})\) For (a): \(E^{\circ}_a = 1.23\,\text{V} - 0.54\,\text{V} = 0.69\,\text{V}\) For (b): \(E^{\circ}_b = -0.14\,\text{V} - 0\,\text{V} = -0.14\,\text{V}\)

Step 4: Determine the standard reduction potentials for (c)

For (c), we have given half-cells, so we can directly look up their standard reduction potentials: \(E_{red,c}^\circ(Ag^+) = 0.80\,\text{V}\) \(E_{red,c}^\circ(AuCl_4^-) = 0.93\,\text{V}\)

Step 5: Calculate the standard cell potential for (c)

Following the same formula as in Step 3, we can calculate the standard cell potential for (c): \(E^{\circ}_c = 0.93\,\text{V} - 0.80\,\text{V} = 0.13\,\text{V}\) Now we have calculated the standard cell potentials for all three voltaic cells: (a) \(E^{\circ}_a = 0.69\,\text{V}\) (b) \(E^{\circ}_b = -0.14\,\text{V}\) (c) \(E^{\circ}_c = 0.13\,\text{V}\)

Key Concepts

Voltaic Cells

Voltaic cells, also known as galvanic cells, are devices that convert chemical energy into electrical energy through redox reactions. In a voltaic cell, there are two half-cells connected by a salt bridge or porous disk which allows ion exchange while keeping the solutions of each half-cell separate. Each half-cell contains an electrode and an electrolyte, where the electrode's material and the electrolyte's contents are determined based on the reaction occurring in that half-cell.
The flow of electrons is what drives electrical energy in the cell, moving from the anode (the site of oxidation) to the cathode (the site of reduction) through an external wire, generating a current. The spontaneous redox reaction in the cell provides the energy needed to push electrons from the anode to the cathode.
Voltaic cells are crucial for many electronic devices and are essentially the working principle in batteries that power gadgets ranging from smartphones to cars.

Reduction Potentials

In electrochemistry, reduction potential (also known as standard electrode potential, E°) is the voltage due to a given half-reaction in the reduction direction at standard conditions: 1 M concentration for solutions, 1 atm pressure for gases, and pure solids or liquids for other phases, all at 25°C. It indicates a substance's tendency to gain electrons and thus be reduced.
Reduction potentials are usually determined relative to the standard hydrogen electrode (SHE), which is assigned a potential of 0 V. The higher the reduction potential, the more likely the species will gain electrons and reduce.

• Positive values indicate a strong tendency to be reduced.
• Negative values indicate a weak tendency to be reduced.

Using reduction potentials, one can predict cell potentials in voltaic cells, enabling the calculation of electrical work the cell can perform.

Electrochemistry

Electrochemistry is the branch of chemistry that explores the relationship between chemical reactions and electricity. It plays a vital role in various applications from powering batteries to electroplating metals to producing chemicals.
The core concepts of electrochemistry involve understanding redox reactions—reactions where the transfer of electrons between chemical species occurs. Electrochemistry focuses on reactions in which electron transfer results in a flow of electric current, separating these processes between two half-cells in a voltaic cell.
Electrochemistry is essential for designing sustainable energy solutions and developing technologies such as fuel cells and photovoltaics. It exploits redox reactions to develop processes like corrosion prevention and energy storage, emphasizing safety and efficiency.

Half-Reactions

Half-reactions are the simplified components of a full redox reaction that either involves oxidation or reduction. Each redox reaction can be split into two half-reactions: one representing the oxidation process and the other representing the reduction process.
In a half-reaction, you will see either the gain or loss of electrons. An oxidation half-reaction shows the loss of electrons, whereas a reduction half-reaction shows the gain of electrons. For instance:

• Oxidation half-reaction: \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \)
• Reduction half-reaction: \( \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu(s)} \)

Balancing half-reactions is crucial to ensure mass and charge conservation. They allow simpler calculations of overall cell potential by focusing separately on the energy changes involved in each electron transfer step of the redox process.