Problem 56 - Chemistry The Molecular Nature of Matter

Question

Write the half-reactions and the balanced cell reaction for the following galvanic cells. (a) \(\operatorname{Cd}(s)\left|\mathrm{Cd}^{2+}(a q) \| \mathrm{Au}^{3+}(a q)\right| \mathrm{Au}(s)\) (b) \(\mathrm{Fe}(s)\left|\mathrm{Fe}^{2+}(a q) \| \operatorname{Br}_{2}(a q), \operatorname{Br}^{-}(a q)\right| \operatorname{Pt}(s)\) (c) \(\operatorname{Cr}(s)\left|\mathrm{Cr}^{3+}(a q) \| \mathrm{Cu}^{2+}(a q)\right| \mathrm{Cu}(s)\)

Step-by-Step Solution

Answer

(a) 3Cd(s) + 2Au^{3+}(aq) -> 3Cd^{2+}(aq) + 2Au(s), (b) Fe(s) + Br_{2}(aq) -> Fe^{2+}(aq) + 2Br^{-}(aq), (c) 2Cr(s) + 3Cu^{2+}(aq) -> 2Cr^{3+}(aq) + 3Cu(s).

Step 1: Identify the Half-Reactions for cell (a)

Write down the oxidation and reduction half-reactions for the cell components. For cadmium: Cd(s) -> Cd^{2+}(aq) + 2e^{-}. For gold: Au^{3+}(aq) + 3e^{-} -> Au(s).

Step 2: Balance the Electrons for cell (a)

Balance the electrons transferred in the half-reactions. Multiply the cadmium half-reaction by 3 and the gold half-reaction by 2 to get 6 electrons in both: 3Cd(s) -> 3Cd^{2+}(aq) + 6e^{-}, 2Au^{3+}(aq) + 6e^{-} -> 2Au(s).

Step 3: Write the Balanced Cell Reaction for cell (a)

Combine the half-reactions canceling out the electrons to get the overall balanced equation: 3Cd(s) + 2Au^{3+}(aq) -> 3Cd^{2+}(aq) + 2Au(s).

Step 4: Identify the Half-Reactions for cell (b)

Write down oxidation and reduction half-reactions. For iron: Fe(s) -> Fe^{2+}(aq) + 2e^{-}. For bromine: Br_{2}(aq) + 2e^{-} -> 2Br^{-}(aq).

Step 5: Balance the Electrons for cell (b)

Both half-reactions are already balanced with 2 electrons each. No further modification is needed.

Step 6: Write the Balanced Cell Reaction for cell (b)

Combine the half-reactions canceling out the electrons: Fe(s) + Br_{2}(aq) -> Fe^{2+}(aq) + 2Br^{-}(aq).

Step 7: Identify the Half-Reactions for cell (c)

Write down the oxidation and reduction half-reactions. For chromium: Cr(s) -> Cr^{3+}(aq) + 3e^{-}. For copper: Cu^{2+}(aq) + 2e^{-} -> Cu(s).

Step 8: Balance the Electrons for cell (c)

Balance the electrons transferred in the half-reactions. Multiply the chromium half-reaction by 2 and the copper half-reaction by 3 to get 6 electrons in both: 2Cr(s) -> 2Cr^{3+}(aq) + 6e^{-}, 3Cu^{2+}(aq) + 6e^{-} -> 3Cu(s).

Step 9: Write the Balanced Cell Reaction for cell (c)

Combine the half-reactions canceling out the electrons: 2Cr(s) + 3Cu^{2+}(aq) -> 2Cr^{3+}(aq) + 3Cu(s).

Key Concepts

Redox Reactions

Redox reactions are chemical processes where oxidation and reduction occur simultaneously. During these reactions, electrons are transferred from one substance to another. The substance that loses electrons is said to be oxidized, while the one gaining electrons is reduced. This transfer of electrons leads to changes in the oxidation states of the atoms involved in the reaction.

Oxidation refers to the loss of electrons, and is often associated with the addition of oxygen or the removal of hydrogen. On the other hand, reduction refers to the gain of electrons, usually coupled with the addition of hydrogen or the removal of oxygen. A helpful mnemonic to remember this is 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain of electrons.

In the provided exercise, for example, cadmium (Cd) gets oxidized from its solid metallic form to a 2+ ion, while gold (Au) ions get reduced to solid gold. Understanding the redox processes is crucial for figuring out the electron flow in galvanic cells.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical reactions. Central to electrochemistry are galvanic cells, which are devices capable of converting chemical energy into electrical energy through redox reactions. These cells play a critical role in batteries and various forms of energy storage devices.

A galvanic cell consists of two half-cells, each containing an electrode and an electrolyte. The electrodes (usually metals or inert conductors like platinum) are where the oxidation and reduction reactions occur. It's essential for students to recognize that the anode is the electrode where oxidation happens and the cathode is where reduction takes place. Electrons flow from the anode to the cathode through an external circuit, while ions move within the electrolytes and salt bridge to maintain charge balance.

In our problem examples, the half-reactions occur at separate electrodes—the oxidation of cadmium at the anode and the reduction of gold at the cathode for cell (a) represent the fundamental components of the galvanic cell's operation.

Balancing Chemical Equations

Balancing chemical equations is a necessary skill in chemistry. It involves making sure that the number of atoms for each element is the same on both sides of the reaction. This accounts for the law of conservation of mass, which states that matter cannot be created nor destroyed in a chemical reaction.

To balance a chemical equation, you must adjust the coefficients—the numbers in front of the compounds or elements—so that the number of atoms of each element is equal on both sides of the equation. For redox reactions, special attention must be paid to the transfer of electrons to ensure that the number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction.

In the textbook solutions, we saw an example of balancing by adjusting coefficients to match the number of electrons exchanged. For instance, in cell (a), the Cd half-reaction was multiplied by 3 and the Au half-reaction by 2 to balance the 6 electrons transferred. This careful balancing leads to the correct stoichiometry in the overall cell reaction, which is fundamental for understanding the exact changes taking place during the electrochemical process.