Problem 58 - Chemistry The Molecular Nature of Matter

Question

Write the cell notation for the following galvanic cells. For half-reactions in which all the reactants are in solution or are gases, assume the use of inert platinum electrodes. $$ \text { (a) } \mathrm{Cd}^{2+}(a q)+\mathrm{Fe}(s) \longrightarrow \mathrm{Cd}(s)+\mathrm{Fe}^{2+}(a q) $$ (b) \(\mathrm{NiO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{Ag}(s) \longrightarrow\) $$ \begin{array}{l} \mathrm{Ni}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Ag}^{+}(a q) \\ \text { (c) } \mathrm{Mg}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cd}(s) \end{array} $$

Step-by-Step Solution

Answer

(a) \(\mathrm{Fe}(s) | \mathrm{Fe}^{2+}(aq) || \mathrm{Cd}^{2+}(aq) | \mathrm{Cd}(s)\). (b) \(\mathrm{Pt}(s) | \mathrm{Ag}(s) | \mathrm{Ag}^{+}(aq) || \mathrm{H}^{+}(aq) | \mathrm{NiO}_{2}(s) | \mathrm{Ni}^{2+}(aq) | \mathrm{Pt}(s)\). (c) \(\mathrm{Mg}(s) | \mathrm{Mg}^{2+}(aq) || \mathrm{Cd}^{2+}(aq) | \mathrm{Cd}(s)\).

Step 1: Identify the Oxidation and Reduction Half-Reactions for (a)

For the reaction \(\mathrm{Cd}^{2+}(aq)+\mathrm{Fe}(s) \longrightarrow \mathrm{Cd}(s)+\mathrm{Fe}^{2+}(aq)\), identify the oxidation and reduction half-reactions. The oxidation half-reaction is where the iron loses electrons: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(aq) + 2e^{-}\). The reduction half-reaction is where cadmium gains electrons: \(\mathrm{Cd}^{2+}(aq) + 2e^{-} \longrightarrow \mathrm{Cd}(s)\).

Step 2: Write the Cell Notation for (a)

The cell notation for the reaction is written as anode | anode solution || cathode solution | cathode. For (a) the anode is iron and the cathode is cadmium. The cell notation is: \(\mathrm{Fe}(s) | \mathrm{Fe}^{2+}(aq) || \mathrm{Cd}^{2+}(aq) | \mathrm{Cd}(s)\).

Step 3: Determine the Half-Reactions for (b)

For the reaction \(\mathrm{NiO}_{2}(s)+4 \mathrm{H}^{+}(aq)+2 \mathrm{Ag}(s) \longrightarrow \mathrm{Ni}^{2+}(aq)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Ag}^{+}(aq)\), the oxidation half-reaction has silver turning into silver ions with the release of electrons: \(2 \mathrm{Ag}(s) \longrightarrow 2 \mathrm{Ag}^{+}(aq) + 2e^{-}\). The reduction half-reaction involves \(\mathrm{NiO}_{2}(s)\) and \(\mathrm{H}^{+}(aq)\) being reduced to \(\mathrm{Ni}^{2+}(aq)\) and water.

Step 4: Write the Cell Notation for (b)

In case (b), we use inert platinum electrodes where reactants or products are in solution or gas form. The cell notation for (b) is: \(\mathrm{Pt}(s) | \mathrm{Ag}(s) | \mathrm{Ag}^{+}(aq) || \mathrm{H}^{+}(aq) | \mathrm{NiO}_{2}(s) | \mathrm{Ni}^{2+}(aq) | \mathrm{Pt}(s)\).

Step 5: Identify the Oxidation and Reduction Half-Reactions for (c)

For the reaction \(\mathrm{Mg}(s) + \mathrm{Cd}^{2+}(aq) \longrightarrow \mathrm{Mg}^{2+}(aq) + \mathrm{Cd}(s)\), we identify the oxidation half-reaction where magnesium loses electrons: \(\mathrm{Mg}(s) \longrightarrow \mathrm{Mg}^{2+}(aq) + 2e^{-}\) and the reduction half-reaction where cadmium ions gain electrons: \(\mathrm{Cd}^{2+}(aq) + 2e^{-} \longrightarrow \mathrm{Cd}(s)\).

Step 6: Write the Cell Notation for (c)

The cell notation for (c) follows the format of anode | anode solution || cathode solution | cathode, thus: \(\mathrm{Mg}(s) | \mathrm{Mg}^{2+}(aq) || \mathrm{Cd}^{2+}(aq) | \mathrm{Cd}(s)\).

Key Concepts

Oxidation Half-Reaction

Oxidation half-reactions are fundamental to understanding electrochemistry. They describe the process where an element loses electrons, which is a characteristic change during a redox reaction. This loss of electrons increases the oxidation state of the element. For example, in the galvanic cell reaction involving magnesium and cadmium, \(\text{Mg}(s) \longrightarrow \text{Mg}^{2+}(aq) + 2e^{-}\), magnesium is undergoing oxidation as it releases two electrons and transforms from a solid metal into its ionic form in solution. Identifying oxidation half-reactions is the first step in analyzing and constructing galvanic cells, as these reactions occur at the anode of the cell.

Reduction Half-Reaction

The reduction half-reaction, in contrast, involves the gain of electrons by a chemical species. When a substance undergoes reduction, its oxidation state decreases. Considering \(\text{Cd}^{2+}(aq) + 2e^{-} \longrightarrow \text{Cd}(s)\), cadmium ions in solution are gaining electrons and being reduced to solid cadmium. This occurs at the cathode in a galvanic cell. Reduction half-reactions, paired with oxidation half-reactions, help in determining the flow of electrons in an electrochemical cell, thereby establishing the direction of the electric current.

Electrochemistry

Electrochemistry is the study of chemical processes that cause electrons to move. This movement of electrons constitutes an electric current that can be harnessed for various applications. The core of electrochemistry is the redox reaction, where oxidation and reduction half-reactions occur. In a galvanic cell, these half-reactions take place in separate compartments, allowing for the spontaneous flow of electrons from the anode to the cathode, generating an electric current. Understanding how these reactions work and are represented, such as \(\text{Fe}(s) | \text{Fe}^{2+}(aq) || \text{Cd}^{2+}(aq) | \text{Cd}(s)\), provides valuable insight into developing batteries, preventing corrosion, and other important applications of electrochemistry.

Inert Platinum Electrodes

Inert platinum electrodes are used in galvanic cells when the reactants or products are in solution or gas form. Platinum is chemically inert, meaning it does not participate in the redox reactions taking place in the cell. It provides a surface for the reactions to occur and acts as a conduit for electrons. In the cell notation for reaction \(\text{b}\), \(\text{Pt}(s) | \text{Ag}(s) | \text{Ag}^{+}(aq) || \text{H}^{+}(aq) | \text{NiO}_{2}(s) | \text{Ni}^{2+}(aq) | \text{Pt}(s)\), platinum electrodes are positioned at the ends of the cell notation, symbolizing their role as the physical structure that completes the circuit without undergoing chemical change.