Question

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: (a) \(\mathrm{C}_{2}(\mathrm{~g})+2 \mathrm{I}^{\mathrm{a}}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s)\) (b) \(\mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q)\) (c) \(\mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q)\) (d) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{NO}(g)+\) \(4 \mathrm{H}_{2} \mathrm{O}(I)+3 \mathrm{Cu}^{2+}(a q)\)

Short answer

Short Answer: The standard emfs for the given reactions are: (a) \(E_{cell} = 0.82\,\mathrm{V}\) (b) \(E_{cell} = 1.86\,\mathrm{V}\) (c) \(E_{cell} = 1.21\,\mathrm{V}\) (d) \(E_{cell} = -0.63\,\mathrm{V}\)

Step-by-step solution

Identify the half-reactions

Write the half-reactions. 1. Oxidation half-reaction: \(2I^{-}(aq) \longrightarrow I_{2}(s) + 2e^{-}\) 2. Reduction half-reaction: \(Cl_{2}(g) + 2e^{-} \longrightarrow 2Cl^{-}(aq)\)

Look up the standard reduction potentials

Find the standard reduction potentials for both half-reactions. According to Appendix E: 1. \(I_{2}(s) + 2e^{-} \longrightarrow 2I^{-}(aq)\) : \(E^\circ_{red,I} = 0.54\,\mathrm{V}\) 2. \(Cl_{2}(g) + 2e^{-} \longrightarrow 2Cl^{-}(aq)\) : \(E^\circ_{red,Cl} = 1.36\,\mathrm{V}\)

Calculate the standard emf

Use the given equation: \(E_{cell} = E_{cathode} - E_{anode} = E^\circ_{red,Cl} - E^\circ_{red,I} = 1.36\,\mathrm{V} - 0.54\,\mathrm{V} = 0.82\,\mathrm{V}\) (b) \(Ni^{2+}(aq) + 2e^{-} \longrightarrow Ni(s)\) \(2Ce^{3+}(aq) + 2e^{-} \longrightarrow 2Ce^{4+}(aq)\)

Identify the half-reactions

Write the half-reactions. 1. Oxidation half-reaction: \(Ni(s) \longrightarrow Ni^{2+}(aq) + 2e^{-}\) 2. Reduction half-reaction: \(2Ce^{4+}(aq) + 2e^{-} \longrightarrow 2Ce^{3+}(aq)\)

Look up the standard reduction potentials

Find the standard reduction potentials for both half-reactions. According to Appendix E: 1. \(Ni^{2+}(aq) + 2e^{-} \longrightarrow Ni(s)\) : \(E^\circ_{red,Ni} = -0.25\,\mathrm{V}\) 2. \(Ce^{4+}(aq) + e^{-} \longrightarrow Ce^{3+}(aq)\) : \(E^\circ_{red,Ce} = 1.61\,\mathrm{V}\)

Calculate the standard emf

Use the given equation: \(E_{cell} = E_{cathode} - E_{anode} = E^\circ_{red,Ce} - E^\circ_{red,Ni} = 1.61\,\mathrm{V} - (-0.25\,\mathrm{V}) = 1.86\,\mathrm{V}\) (c) \(Fe^{2+}(aq) + 2e^{-} \longrightarrow Fe(s)\) \(Fe^{3+}(aq) + e^{-} \longrightarrow Fe^{2+}(aq)\)

Identify the half-reactions

Write the half-reactions. 1. Oxidation half-reaction: \(Fe(s) \longrightarrow Fe^{2+}(aq) + 2e^{-}\) 2. Reduction half-reaction: \(2Fe^{3+}(aq) + 2e^{-} \longrightarrow 2Fe^{2+}(aq)\)

Look up the standard reduction potentials

Find the standard reduction potentials for both half-reactions. According to Appendix E: 1. \(Fe^{2+}(aq) + 2e^{-} \longrightarrow Fe(s)\) : \(E^\circ_{red,Fe2} = -0.44\,\mathrm{V}\) 2. \(Fe^{3+}(aq) + e^{-} \longrightarrow Fe^{2+}(aq)\) : \(E^\circ_{red,Fe3} = 0.77\,\mathrm{V}\)

Calculate the standard emf

Use the given equation: \(E_{cell} = E_{cathode} - E_{anode} = E^\circ_{red,Fe3} - E^\circ_{red,Fe2} = 0.77\,\mathrm{V} - (-0.44\,\mathrm{V}) = 1.21\,\mathrm{V}\) For reaction (d), we will need to split it into three half-reactions due to the presence of multiple elements changing oxidation states. (d) \(NO_{3}^{-}(aq) + 2H^{+}(aq) + e^{-} \longrightarrow NO(g) + H_{2}O(l)\) \(2H^{+}(aq) + 2e^{-} \longrightarrow H_{2}(g)\) \(Cu^{2+}(aq) + 2e^{-} \longrightarrow Cu(s)\)

Identify the half-reactions

Write the half-reactions. 1. Oxidation half-reaction: \(2NO(g) + H_{2}O(l) \longrightarrow 2NO_{3}^{-}(aq) + 2H^{+}(aq) + 2e^{-}\) 2. Reduction half-reaction: \(2H^{+}(aq) + 2e^{-} \longrightarrow H_{2}(g)\) 3. Reduction half-reaction: \(3Cu(s) \longrightarrow 3Cu^{2+}(aq) + 6e^{-}\)

Balance the electrons in the half-reactions

Before looking up the standard reduction potentials, balance the electrons in the half-reactions. 1. The NO half-reaction releases 2 electrons. 2. The H_{2} half-reaction consumes 2 electrons. 3. The Cu half-reaction consumes 6 electrons. The net release of electrons from the oxidation half-reaction should equal the net consumption of electrons by the reduction half-reactions. Therefore, multiply the Cu half-reaction by 2/3 to make it consume 4 electrons: \(2Cu(s) \longrightarrow 2Cu^{2+}(aq) + 4e^{-}\)

Look up the standard reduction potentials

Find the standard reduction potentials for all balanced half-reactions. According to Appendix E: 1. \(NO_{3}^{-}(aq) + 2H^{+}(aq) + e^{-} \longrightarrow NO(g) + H_{2}O(l)\) : \(E^\circ_{red,NO} = 0.97\,\mathrm{V}\) 2. \(2H^{+}(aq) + 2e^{-} \longrightarrow H_{2}(g)\) : \(E^\circ_{red,H} = 0\,\mathrm{V}\) 3. \(Cu^{2+}(aq) + 2e^{-} \longrightarrow Cu(s)\) : \(E^\circ_{red,Cu} = 0.34\,\mathrm{V}\)

Calculate the standard emf

Use the given equation: \(E_{cell} = E_{cathode1} - E_{anode} + E_{cathode2} = E^\circ_{red,H} - E^\circ_{red,NO} + E^\circ_{red,Cu} = 0\,\mathrm{V} - 0.97\,\mathrm{V} + 0.34\,\mathrm{V} = -0.63\,\mathrm{V}\)

Key concepts

Standard Reduction Potentials

Standard reduction potentials are critical in electrochemistry. They represent the tendency of a chemical species to acquire electrons and be reduced. These potentials are measured under standard conditions: 25°C, 1 atm pressure, and 1 M concentration for all aqueous species. In electrochemical reactions, such potentials are listed as reduction potentials and are used to predict the direction of redox reactions.

Each half-reaction has its own standard reduction potential, denoted as \(E^0_{red}\). For instance, in the exercise, the reduction half-reaction of \(Cl_2(g)\) converting into \(2Cl^-(aq)\) has a reduction potential \(E^0_{red,Cl} = 1.36\,\mathrm{V}\).

Higher positive values indicate a greater likelihood of a substance being reduced, as seen in the comparison of \(Cl_2\) and \(I_2\), where chlorine is more readily reduced.

Electromotive Force Calculation

Electromotive force (emf) is the driving force behind the flow of electrons in an electrochemical cell. To calculate the standard emf \(E_{cell}\) for a given reaction, we utilize standard reduction potentials. The formula is quite straightforward:

\[ E_{cell} = E_{cathode} - E_{anode} \]

In the context of a redox reaction, the cathode is the site of reduction, while the anode is the site of oxidation. Using the standard reduction potentials from the example, where \(E_{cathode}\) for chloride is 1.36 V and \(E_{anode}\) for iodide is 0.54 V, the emf is calculated as:

\[ E_{cell} = 1.36\,\mathrm{V} - 0.54\,\mathrm{V} = 0.82\,\mathrm{V} \]

This positive result indicates a spontaneous reaction under standard conditions.

Oxidation and Reduction Half-Reactions

Understanding half-reactions is essential for grasping redox processes. A reaction consists of two parts: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. These are better understood as half-reactions.

In the provided exercise, the reaction \(2I^-(aq)\) oxidizing to \(I_2(s)\) represents the oxidation half-reaction, while \(Cl_2(g)\) reducing to \(2Cl^-(aq)\) is the reduction half-reaction. Oxidation involves electron loss, and that is why the \(I^-\) species loses electrons to form \(I_2\).

Always identify the species undergoing oxidation and reduction separately to analyze redox reactions correctly. This separation allows for the correct application of the Nernst equation or other electrochemical principles.

Balancing Redox Reactions

Balancing redox reactions ensures the conservation of mass and charge, crucial in achieving accurate chemical equations. It involves a methodical approach:

• Separate the reaction into its oxidation and reduction half-reactions.
• Balance the number of atoms other than hydrogen and oxygen.
• Balance oxygen atoms by adding \(H_2O\).
• Balance hydrogen atoms by adding \(H^+\).
• Balance the charge by adding electrons.
• Finally, equalize the electron transfer between the half-reactions.

This process was evident when adjusting the copper reaction in the given exercise. By balancing the electrons lost and gained, reactants and products appeared in correct proportions, aligning with laws of conservation. Accurate balancing allows correct emf calculations, ensuring predictive reactions in galvanic or voltaic cells.