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) \(2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ca}(s) \longrightarrow 2 \mathrm{Al}(s)+3 \mathrm{Ca}^{2+}(a q)\)

Short answer

The overall standard emfs for the given reactions are: (a) \(1.90\, V\) (b) \(1.97\, V\) (c) \(1.21\, V\) (d) \(1.21\, V\)

Step-by-step solution

(Reaction A) Identify oxidation and reduction half-reactions

For reaction A, we have: Oxidation: \(\mathrm{2I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s) + 2e^-\) Reduction: \(\mathrm{Cl}_{2}(g) + 2e^- \longrightarrow 2\mathrm{Cl}^{-}(a q)\)

(Reaction A) Look up standard potentials

According to Appendix E: Oxidation: \(E^{\circ}_{\mathrm{I}^{-}/\mathrm{I}_{2}}=-0.54\, V\) Reduction: \(E^{\circ}_{\mathrm{Cl}_{2}/\mathrm{Cl}^{-}}=+1.36\, V\)

(Reaction A) Reverse the potential of oxidation half-reaction

The oxidation half-reaction needs to be reversed: \(E^{\circ}_{\mathrm{I}_{2}/\mathrm{I}^{-}} = -E^{\circ}_{\mathrm{I}^{-}/\mathrm{I}_{2}}=0.54\, V\)

(Reaction A) Calculate overall emf

Finally, we combine the half-reactions: \(E^{\circ}_{\text{cell}}=E^{\circ}_{\mathrm{I}_{2}/\mathrm{I}^{-}}+E^{\circ}_{\mathrm{Cl}_{2}/\mathrm{Cl}^{-}}=0.54\, V+1.36\, V=1.90\, V\) Reaction A's overall standard emf is \(1.90\, V\). Similar steps will be performed for the remaining reactions:

(Reaction B) Steps 1-4

1. Oxidation: \(\mathrm{Ni}(s) \longrightarrow \mathrm{Ni}^{2+}(a q) + 2e^-\) Reduction: \(\mathrm{2Ce}^{4+}(a q) + 2e^- \longrightarrow 2\mathrm{Ce}^{3+}(a q)\) 2. Oxidation: \(E^{\circ}_{\mathrm{Ni}^{2+}/\mathrm{Ni}}=-0.25\, V\) Reduction: \(E^{\circ}_{\mathrm{Ce}^{4+}/\mathrm{Ce}^{3+}}=+1.72\, V\) 3. \(E^{\circ}_{\mathrm{Ni}/\mathrm{Ni}^{2+}} = 0.25\, V\) 4. \(E^{\circ}_{\text{cell}}=0.25\, V+1.72\, V=1.97\, V\) Reaction B's overall standard emf is \(1.97\, V\).

(Reaction C) Steps 1-4

1. Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q) + 2e^-\) Reduction: \(\mathrm{2Fe}^{3+}(a q) + 2e^- \longrightarrow 2\mathrm{Fe}^{2+}(a q)\) 2. Oxidation: \(E^{\circ}_{\mathrm{Fe}^{2+}/\mathrm{Fe}}=-0.44\, V\) Reduction: \(E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}=+0.77\, V\) 3. \(E^{\circ}_{\mathrm{Fe}/\mathrm{Fe}^{2+}} = 0.44\, V\) 4. \(E^{\circ}_{\text{cell}}=0.44\, V+0.77\, V=1.21\, V\) Reaction C's overall standard emf is \(1.21\, V\).

(Reaction D) Steps 1-4

1. Oxidation: \(\mathrm{3Ca}(s) \longrightarrow 3\mathrm{Ca}^{2+}(a q) + 6e^-\) Reduction: \(\mathrm{2Al}^{3+}(a q) + 6e^- \longrightarrow 2\mathrm{Al}(s)\) 2. Oxidation: \(E^{\circ}_{\mathrm{Ca}^{2+}/\mathrm{Ca}}=-2.87\, V\) Reduction: \(E^{\circ}_{\mathrm{Al}^{3+}/\mathrm{Al}}=-1.66\, V\) 3. \(E^{\circ}_{\mathrm{Ca}/\mathrm{Ca}^{2+}} = 2.87\, V\) 4. \(E^{\circ}_{\text{cell}}=2.87\, V+(-1.66\, V)=1.21\, V\) Reaction D's overall standard emf is \(1.21\, V\).

Key concepts

Electrochemistry

Electrochemistry is a fascinating branch of chemistry that deals with the interaction between electrical energy and chemical change. It helps us understand how reactions can either produce electricity or occur because of electricity. One of the main applications of electrochemistry is in batteries, where chemical reactions are used to produce electrical energy.

In the context of electrochemistry, we deal with electrodes immersed in an electrolyte solution, completing a circuit that can drive a chemical reaction or generate a flow of electrons. This concept is essential for understanding various phenomena such as corrosion, electroplating, and many other processes important in everyday life and industrial applications.

• Electrodes: Conductive materials that allow electrons to enter or exit the system.
• Electrolyte: A chemical substance that produces an electrically conducting solution when dissolved.
• Electrochemical cell: A device that generates electrical energy from a chemical reaction or facilitates a chemical reaction through the introduction of electrical energy.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, also known as redox reactions, are chemical reactions involving the transfer of electrons between two species. These reactions are characterized by changes in the oxidation states of reactants. In these reactions, one substance loses electrons (oxidation) and the other gains electrons (reduction).

The substance that donates electrons is called the reducing agent, while the one that accepts electrons is called the oxidizing agent. Redox reactions are crucial in many biological and chemical processes, including respiration, photosynthesis, and combustion.

• Oxidation: Loss of electrons or an increase in oxidation state by a molecule, atom, or ion.
• Reduction: Gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.
• Oxidizing agent: The reactant that gains electrons in a redox reaction.
• Reducing agent: The reactant that loses electrons in a redox reaction.

Standard Electromotive Force (emf)

The standard electromotive force (emf) of a reaction is a measure of the energy released or consumed when an electrochemical reaction occurs. It is expressed in volts (V) and can be calculated by combining the standard electrode potentials of the reaction's half-reactions.

The standard emf can be used to predict the direction of the reaction and its spontaneity under standard conditions (1 M concentration, 1 atm pressure, and 25°C temperature). A positive standard emf indicates a spontaneous reaction, while a negative one suggests that the reaction is non-spontaneous under standard conditions.

• Standard conditions: 1 M concentration, 1 atm pressure, 25°C temperature.
• Spontaneity: A positive emf means the reaction can occur naturally without added energy.
• Volts (V): The SI unit of electric potential or electromotive force.

Half-Reaction Method

The half-reaction method is a systematic approach to balancing redox reactions. This method separates the oxidation and reduction processes, allowing for the individual balancing of each half-reaction before combining them into the overall balanced equation.

To balance a redox reaction using the half-reaction method, start by dividing the reaction into its oxidation and reduction components. Each half-reaction is balanced separately with respect to mass and charge, usually by adjusting electron counts and adding ions to balance the changes.

• Identify oxidation and reduction half-reactions.
• Balance each half-reaction for mass and charge.
• Combine the half-reactions, ensuring that the electrons lost and gained are equal.

Using this method provides an efficient, step-by-step process that enhances clarity and understanding when solving electrochemical problems.