Problem 37 - Chemistry: The Central Science in SI Units, Global Edition

Question

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

Step-by-Step Solution

Answer

The standard emf values for the given reactions are: (a) 0.82 V (b) -1.46 V (c) 1.21 V (d) 0.62 V

Step 1: Half-Reactions and Standard Reduction Potentials for Reaction (a)

For this reaction, we have: Oxidation half-reaction: 2 I⁻(aq) → I₂(s) + 2 e⁻ Reduction half-reaction: Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) From Appendix E: E°(I⁻/I₂) = +0.54 V E°(Cl₂/Cl⁻) = +1.36 V #Step 2: Calculate the Standard emf#

Step 2: Standard emf for Reaction (a)

Using the Nernst equation: \[E°_{cell} = E°_{cathode} - E°_{anode}\] For this reaction, emf is: \[E°_{cell} = E°(Cl₂/Cl⁻) - E°(I⁻/I₂) = 1.36 - 0.54 = 0.82 \ V\] #b) Reaction: Ni(s) + 2 Ce⁴⁺(aq) → Ni²⁺(aq) + 2 Ce³⁺(aq)# #Step 1: Identify Half-Reactions and Standard Reduction Potentials#

Step 3: Half-Reactions and Standard Reduction Potentials for Reaction (b)

For this reaction, we have: Oxidation half-reaction: Ni(s) → Ni²⁺(aq) + 2 e⁻ Reduction half-reaction: 2 Ce⁴⁺(aq) + 2 e⁻ → 2 Ce³⁺(aq) From Appendix E: E°(Ni²⁺/Ni) = -0.26 V E°(Ce⁴⁺/Ce³⁺) = -1.72 V #Step 2: Calculate the Standard emf#

Step 4: Standard emf for Reaction (b)

Using the Nernst equation, the standard emf is: \[E°_{cell} = E°(Ce⁴⁺/Ce³⁺) - E°(Ni²⁺/Ni) = -1.72 + 0.26 = -1.46 \ V\] #c) Reaction: Fe(s) + 2 Fe³⁺(aq) → 3 Fe²⁺(aq)# #Step 1: Identify Half-Reactions and Standard Reduction Potentials#

Step 5: Half-Reactions and Standard Reduction Potentials for Reaction (c)

For this reaction, we have: Oxidation half-reaction: Fe(s) → Fe²⁺(aq) + 2 e⁻ Reduction half-reaction: 2 Fe³⁺(aq) + 2 e⁻ → 2 Fe²⁺(aq) From Appendix E: E°(Fe²⁺/Fe) = -0.44 V E°(Fe³⁺/Fe²⁺) = +0.77 V #Step 2: Calculate the Standard emf#

Step 6: Standard emf for Reaction (c)

Using the Nernst equation, the standard emf is: \[E°_{cell} = E°(Fe³⁺/Fe²⁺) - E°(Fe²⁺/Fe) = 0.77 + 0.44 = 1.21 \ V\] #d) Reaction: 2 NO₃⁻(aq) + 8 H⁺(aq) + 3 Cu(s) → 2 NO(g) + 4 H₂O(l) + 3 Cu²⁺(aq)# #Step 1: Identify Half-Reactions and Standard Reduction Potentials#

Step 7: Half-Reactions and Standard Reduction Potentials for Reaction (d)

For this reaction, we have: Oxidation half-reaction: 3 Cu(s) → 3 Cu²⁺(aq) + 6 e⁻ Reduction half-reaction: 2 NO₃⁻(aq) + 8 H⁺(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O(l) From Appendix E: E°(Cu²⁺/Cu) = +0.34 V E°(NO₃⁻/NO) = +0.96 V #Step 2: Calculate the Standard emf#

Step 8: Standard emf for Reaction (d)

Using the Nernst equation, the standard emf is: \[E°_{cell} = E°(NO₃⁻/NO) - E°(Cu²⁺/Cu) = 0.96 - 0.34 = 0.62 \ V\] In summary, the standard emf values for the given reactions are: (a) 0.82 V (b) -1.46 V (c) 1.21 V (d) 0.62 V

Key Concepts

Standard Reduction Potentials

In electrochemistry, standard reduction potentials are essential for understanding how different elements and compounds behave in redox reactions. A standard reduction potential, denoted as \( E^0 \), is the tendency of a chemical species to acquire electrons and thereby be reduced. These potentials are measured under standard conditions of 1 M concentration, 1 atm pressure, and at a temperature of 25°C.Standard reduction potentials are instrumental for predicting the direction of electron flow in redox reactions. They provide a measure of how much a particular substance wants to gain electrons compared to a reference cell, typically the standard hydrogen electrode, which has a reduction potential of 0 V.Understanding these potentials allows us to determine which species in a reaction is more likely to gain electrons (be reduced) and which one will lose electrons (be oxidized). By comparing the standard reduction potential of different half-reactions, we can predict how reactions will occur spontaneously. This is because reactions progress in a direction that results in the formation of the most stable products, typically those formed from species with higher reduction potentials.

EMF Calculation

The electromotive force (emf) of a cell is an important aspect of electrochemical reactions, indicating the voltage or potential difference between two electrodes when no current is flowing. The standard emf is symbolized as \( E^0_{cell} \) and is calculated using the standard reduction potentials of the cathode and anode reactions in a galvanic cell.For the calculation of the standard emf, we use the formula:\[E^0_{cell} = E^0_{cathode} - E^0_{anode}\]This equation allows us to subtract the standard reduction potential of the oxidation reaction (anode) from that of the reduction reaction (cathode). It's essential to remember that the cathode is where reduction happens, and it has a more positive potential than the anode, where oxidation occurs.By calculating the standard emf, we can predict whether a particular electrochemical reaction is spontaneous. A positive \( E^0_{cell} \) indicates a spontaneous reaction under standard conditions, while a negative value suggests non-spontaneity.

Half-Reactions

In any redox reaction, the overall process can be broken down into two simultaneous reactions called half-reactions. These half-reactions include the oxidation half-reaction, where a species loses electrons, and the reduction half-reaction, where a species gains electrons. Identifying half-reactions is crucial for calculating standard emf, as each half-reaction has its own standard reduction potential. To determine the overall cell reaction, one must:

• Identify the oxidation half-reaction by determining which species loses electrons.
• Identify the reduction half-reaction by specifying which species gains electrons.
• Balance both half-reactions for mass and charge, ensuring electrons lost in the oxidation half-reaction are gained in the reduction half-reaction.

Utilizing half-reactions allows chemists to understand the electron transfer process within a cell and enables calculation of cell potentials, making it a cornerstone concept in the study of electrochemistry.