Problem 40 - Chemistry Matter and Change

Question

Write the balanced chemical equation for the standard cell notations listed below. $$ \begin{array}{l}{\text { a. } \mathrm{I}^{-}\left|\mathrm{I}_{2} \| \mathrm{Fe}^{3+}\right| \mathrm{Fe}^{2+}} \\ {\text { b. } \mathrm{sn}\left|\mathrm{Sn}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}} \\\ {\text { c. } \mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cd}^{2+}\right| \mathrm{Cd}}\end{array} $$

Step-by-Step Solution

Answer

a. \( 2 I^- + 2 Fe^{3+} \rightarrow I_2 + 2 Fe^{2+} \) b. \( Sn + 2 Ag^+ \rightarrow Sn^{2+} + 2 Ag \) c. \( Zn + Cd^{2+} \rightarrow Zn^{2+} + Cd \)

Step 1: Identify the half-reactions

For each cell notation, we can identify the species undergoing reduction and oxidation. The species on the left side of the notation undergoes oxidation (loses electrons), and the species on the right side undergoes reduction (gains electrons). a. I⁻ | I2 || Fe³⁺ | Fe²⁺ Oxidation: I⁻ → I2 Reduction: Fe³⁺ → Fe²⁺ b. Sn | Sn²⁺ || Ag⁺ | Ag Oxidation: Sn → Sn²⁺ Reduction: Ag⁺ → Ag c. Zn | Zn²⁺ || Cd²⁺ | Cd Oxidation: Zn → Zn²⁺ Reduction: Cd²⁺ → Cd

Step 2: Balance the half-reactions

Next, balance the half-reactions by balancing the atoms, charges, and then adding enough electrons to the appropriate side to ensure charge balance. a. Oxidation: 2 I⁻ → I2 + 2e⁻ Reduction: Fe³⁺ + e⁻ → Fe²⁺ b. Oxidation: Sn → Sn²⁺ + 2e⁻ Reduction: Ag⁺ + e⁻ → Ag c. Oxidation: Zn → Zn²⁺ + 2e⁻ Reduction: Cd²⁺ + 2e⁻ → Cd

Step 3: Combine the half-reactions

Combine the balanced half-reactions for each cell, making sure the number of electrons is the same in both half-reactions. If necessary, multiply one or both half-reactions by an appropriate factor to equalize the number of electrons. a. 2 I⁻ → I2 + 2e⁻ 2(Fe³⁺ + e⁻ → Fe²⁺) Overall: 2 I⁻ + 2 Fe³⁺ → I2 + 2 Fe²⁺ b. Sn → Sn²⁺ + 2e⁻ 2(Ag⁺ + e⁻ → Ag) Overall: Sn + 2 Ag⁺ → Sn²⁺ + 2 Ag c. Zn → Zn²⁺ + 2e⁻ Cd²⁺ + 2e⁻ → Cd Overall: Zn + Cd²⁺ → Zn²⁺ + Cd Now we have the balanced chemical equations for each standard cell notation.

Key Concepts

Oxidation and Reduction

In chemistry, oxidation and reduction reactions are processes that involve the exchange of electrons between two species. These reactions occur simultaneously; when one substance is oxidized, another is reduced.
Oxidation refers to the loss of electrons by a molecule, atom, or ion. For instance, in the reaction \( ext{I}^- \rightarrow ext{I}_2 \), iodide ions (\( ext{I}^- \)) lose electrons and transform into iodine molecules (\( ext{I}_2 \)).
Reduction, on the other hand, is the gain of electrons. For example, in the process \( ext{Fe}^{3+} \rightarrow ext{Fe}^{2+} \), iron(III) ions gain electrons, becoming iron(II) ions. Thus, one cannot happen without the other. In the shorthand notation of electrochemical cells, the left side is typically for oxidation, while the right is for reduction.

Half-Reactions

Half-reactions break down a redox reaction into its oxidation and reduction parts. They make it easier to analyze each process separately and understand how electrons are transferred between species.
For example, let's consider the electrochemical cell notation of \( ext{Sn} | ext{Sn}^{2+} || ext{Ag}^+ | ext{Ag} \). The oxidation half-reaction is \( ext{Sn} \rightarrow ext{Sn}^{2+} + 2e^- \), showing that tin loses electrons. The reduction half-reaction would be \( ext{Ag}^+ + e^- \rightarrow ext{Ag} \), where silver ions gain electrons.
By balancing the electrons in each half-reaction, we ensure that the number of electrons lost in oxidation matches the number of electrons gained in reduction, maintaining the principle of charge conservation.

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa through redox reactions. These cells consist of two electrodes placed in separate compartments known as half-cells, connected by an electrolyte bridge allowing ions to move.
The cell is often represented by standard cell notations, like \( ext{Zn} | ext{Zn}^{2+} || ext{Cd}^{2+} | ext{Cd} \). Here, each "|" separates different phases or entities: metals from their ions, and the double line "||" represents the salt bridge.
This system facilitates the flow of electrons through an external circuit from the anode (oxidation site) to the cathode (reduction site), enabling cell operation and energy conversion.

Electron Transfer Balancing

Balancing electron transfer is crucial in redox reactions to ensure that the number of electrons lost equals the number gained, preserving charge neutrality.
To achieve this, we may need to adjust the coefficients of the half-reactions. For instance, if an oxidation half-reaction produces two electrons \( (e.g., \; 2 ext{e}^- \)) and the reduction half only involves one electron \( ( ext{e.g.,} \; ext{Fe}^{3+} + ext{e}^- \rightarrow ext{Fe}^{2+}) \), we multiply the reduction reaction by two, making it \( 2 ext{Fe}^{3+} + 2 ext{e}^- \rightarrow 2 ext{Fe}^{2+} \).
This principle allows us to construct fully balanced chemical equations from the half-reactions by aligning electron numbers, like \( 2 ext{I}^- + 2 ext{Fe}^{3+} \rightarrow ext{I}_2 + 2 ext{Fe}^{2+} \), maintaining consistency with the core idea of electron conservation.