Question

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

Short answer

The standard emf for the given reactions are: (a) +0.82 V, (b) +1.84 V, (c) +1.21 V, and (d) +1.30 V.

Step-by-step solution

(a) Identify half-reactions

For the reaction \[ \text{Cl}_2(g) + 2\text{I}^-(aq) \longrightarrow 2\text{Cl}^-(aq) + \text{I}_2(s) \] the half-reactions are: Reduction: \(\text{Cl}_2(g) + 2\text{e}^- \longrightarrow 2\text{Cl}^-(aq)\) Oxidation: \(2\text{I}^-(aq) \longrightarrow \text{I}_2(s) + 2\text{e}^-\)

(a) Look up standard reduction potentials

Consult Appendix E for the standard reduction potentials of the half-reactions: Reduction (Cl2): \(E_{cathode}^° = +1.36 V\) Oxidation (I-): \(E_{anode}^° = +0.54 V\) (note: reverse the sign as it is undergoing oxidation)

(a) Calculate the standard emf

Now, use the formula to calculate the standard emf: \[ E_{cell}^° = E_{cathode}^° - E_{anode}^° = (+1.36 V) - (+0.54 V) = +0.82 V \]

(b) Identify half-reactions

For the reaction \[ \text{Ni}(s) + 2\text{Ce}^{4+}(aq) \longrightarrow \text{Ni}^{2+}(aq) + 2\text{Ce}^{3+}(aq) \] the half-reactions are: Reduction: \(\text{Ce}^{4+}(aq) + \text{e}^- \longrightarrow \text{Ce}^{3+}(aq)\) Oxidation: \(\text{Ni}(s) \longrightarrow \text{Ni}^{2+}(aq) + 2\text{e}^-\)

(b) Look up standard reduction potentials

Consult Appendix E for the standard reduction potentials of the half-reactions: Reduction (Ce4+): \(E_{cathode}^° = +1.61 V\) Oxidation (Ni): \(E_{anode}^° = -0.23 V\) (note: reverse the sign as it is undergoing oxidation)

(b) Calculate the standard emf

Now, use the formula to calculate the standard emf: \[ E_{cell}^° = E_{cathode}^° - E_{anode}^° = (+1.61 V) - (-0.23 V) = +1.84 V \] Similarly, we will find standard emf for reactions (c) and (d).

(c) Standard emf for the reaction

Following the same procedure as before, we have: \[ E_{cell}^° = (+0.77 V) - (-0.44 V) = +1.21 V \]

(d) Standard emf for the reaction

Following the same procedure as before, we have: \[ E_{cell}^° = (+0.96 V) - (-0.34 V) = +1.30 V \]

Key concepts

Understanding Electrochemistry and Its Importance

Electrochemistry is a fascinating branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It plays a crucial role in various aspects of modern life, from batteries that power our devices to industrial processes that manufacture chemical products. At its core, electrochemistry focuses on redox or oxidation-reduction reactions, where electrons are transferred between atoms or molecules. This transfer of electrons is what generates electrical current in electrochemical cells.

One of the critical principles of electrochemistry is the concept of standard reduction potentials. These are values that measure the tendency of a chemical species to gain electrons and be reduced, usually recorded under standard conditions (1 M concentration, 1 atm pressure, and 25°C). The standard reduction potential is given in volts and is a fundamental parameter in determining the direction and spontaneity of redox reactions.

Calculating Standard EMF (Electromotive Force)

Standard electromotive force (EMF) calculation is essential in predicting how a chemical reaction can produce or consume electrical energy. The standard EMF of a cell represents the net voltage difference between two half-cells when they are connected. To calculate it, you need to know the standard reduction potentials of the anode (oxidation site) and cathode (reduction site). The standard EMF is found using the formula:
\[E_{cell}^° = E_{cathode}^° - E_{anode}^°\]
Here, you must be cautious to reverse the sign of the anode potential, as it pertains to oxidation. By knowing the standard EMF, which is measured in volts, you can predict whether the reaction will occur spontaneously (if positive) and the potential electrical energy that can be harnessed from the reaction.

It is also crucial to ensure that the number of electrons lost in the oxidation reaction equals the number gained in the reduction reaction. This balance is what allows the overall reaction to proceed and is an integral part of solving redox problems in electrochemistry.

Navigating Redox Reactions

Redox reactions are the driving force behind electrochemical cells. They consist of two interrelated processes - reduction, where a substance gains electrons, and oxidation, where a substance loses electrons. These reactions can be broken down into half-reactions, each representing either the reduction or oxidation process. Identifying these half-reactions is a critical step when solving electrochemistry problems.

After identifying the half-reactions, you consult a reference like Appendix E in your textbook to find the standard reduction potentials, crucial for calculating the standard EMF of the entire cell. Remembering that the standard reduction potential of the oxidizing agent needs its sign reversed is a common sticking point that can lead to mistakes in calculations. By mastering redox reactions and their calculations, students gain insights into how chemical energy is converted into electrical energy and vice versa, an understanding that is vital for fields such as energy storage, metallurgy, and environmental science.

By studying examples like the conversion of chlorine gas to chloride ions or the oxidation of nickel metal, learners can appreciate the real-world applications of these principles and thus deepen their understanding of both the theoretical and practical aspects of this dynamic field.