Problem 37 - Chemistry An Atoms First Approach

Question

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)\)

Step-by-Step Solution

Answer

a. Anode: \(Cl_{2}(g) \rightarrow 2Cl^{-}(a q)\); Cathode: \(2Cr^{3+}(a q) \rightarrow Cr_{2} O_{7}^{2-}(a q)\); Galvanic cell: Cl2 bubbled through Cl⁻ solution (anode), Cr³⁺ solution (cathode) connected by a salt bridge; Overall equation: \(Cr^{3+}(a q)+Cl_{2}(g) \rightleftharpoons Cr_{2} O_{7}^{2-}(a q)+Cl^{-}(a q)\) b. Anode: \(Mg(s) \rightarrow Mg^{2+}(a q) + 2e^-\); Cathode: \(Cu^{2+}(a q) + 2e^- \rightarrow Cu(s)\); Galvanic cell: Mg(s) electrode (anode), Cu²⁺ solution and Cu(s) electrode (cathode) connected by a salt bridge; Overall equation: \(Cu^{2+}(a q)+Mg(s) \rightleftharpoons Mg^{2+}(a q)+Cu(s)\)

Step 1: a. Cr3+(aq) + Cl2(g) ⇌ Cr2O72−(aq) + Cl−(aq)

1. Identify the half-reactions: We need to break the overall reaction into reduction and oxidation half-reactions: - Oxidation half-reaction: \(\mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{Cl}^{-}(a q)\) - Reduction half-reaction: \(\mathrm{Cr}^{3+}(a q) \rightarrow \frac{1}{2}\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)\), multiply by 2 to balance the Cr atoms: \(2\mathrm{Cr}^{3+}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)\) 2. Identify the anode and cathode: - Anode (oxidation): \(\mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{Cl}^{-}(a q)\) - Cathode (reduction): \(2\mathrm{Cr}^{3+}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)\) 3. Sketch the galvanic cell: - Anode: Chlorine gas is bubbled through an aqueous solution containing Cl⁻ ions. Electrons flow from the anode to the cathode through an external wire. - Cathode: Chromium(III) ions are present in the aqueous solution to form dichromate ions, Cr2O72-. - Salt bridge: A salt bridge or porous disk connects the two half-cells to maintain charge neutrality. 4. Write the overall balanced equation: The overall balanced equation is given in the problem statement itself: \(\mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\)

Step 2: b. Cu2+(aq) + Mg(s) ⇌ Mg2+(aq) + Cu(s)

1. Identify the half-reactions: We need to break the overall reaction into reduction and oxidation half-reactions: - Oxidation half-reaction: \(\mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(a q) + 2e^-\) - Reduction half-reaction: \(\mathrm{Cu}^{2+}(a q) + 2e^- \rightarrow \mathrm{Cu}(s)\) 2. Identify the anode and cathode: - Anode (oxidation): \(\mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(a q) + 2e^-\) - Cathode (reduction): \(\mathrm{Cu}^{2+}(a q) + 2e^- \rightarrow \mathrm{Cu}(s)\) 3. Sketch the galvanic cell: - Anode: Mg(s) metal electrode is oxidized to Mg²⁺(aq) ions and loses 2 electrons. - Cathode: Cu²⁺(aq) ions are reduced to Cu(s) metal by gaining 2 electrons. - Salt bridge: A salt bridge or porous disk connects the two half-cells to maintain charge neutrality. 4. Write the overall balanced equation: The overall balanced equation is given in the problem statement itself: \(\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)\)

Key Concepts

Electron Flow

In a galvanic cell, electrons flow through an external circuit as a result of redox reactions. The direction of electron movement is pivotal to the operation of the cell. Electrons always flow from the anode, where oxidation occurs, to the cathode, where reduction takes place. This flow is essential for maintaining the charge balance within the cell. The path of electrons can be represented as a line moving from one half-cell to the other, traveling through an external wire. This external path is of utmost importance as it allows for work to be done, such as lighting a bulb.

• Electrons release energy at the anode, generating electrical current.
• At the cathode, electrons are absorbed during the reduction process.
• This flow continues until the reactants are used up or the circuit is broken.

Understanding electron flow is critical for correctly sketching and constructing galvanic cells and applying their concepts practically.

Anode and Cathode Identification

Identifying the anode and cathode is a crucial step in setting up and understanding galvanic cells. The naming is based on the nature of the redox reactions occurring at each electrode. The anode is where oxidation happens — this is the electrode that loses electrons. Conversely, the cathode is the site of reduction — here, electrons are gained. A helpful mnemonic is OIL RIG: Oxidation Is Loss, Reduction Is Gain.

• In a galvanic cell, look for the oxidation reaction to find the anode.
• The reduction reaction will lead you to the cathode.
• Terms of electron movement, electrons flow from anode to cathode naturally.

Successful identification sets the stage for understanding the complete mechanism of galvanic cells.

Half-Reactions

In the study of redox reactions, breaking down the overall reaction into half-reactions is vital for clarity. Half-reactions separately illustrate the oxidation and reduction processes, allowing for easier balancing of electrons. Each half-reaction includes species either gaining or losing electrons, and sometimes, additional steps to balance atoms and charges take place.

• Oxidation half-reaction: In this process, a substance loses electrons, represented with the electrons as products.
• Reduction half-reaction: Here, a material gains electrons, with electrons shown as reactants.
• Half-reactions must balance not only in terms of atoms but also charges by adding electrons appropriately.

Constructing correct half-reactions is central to understanding how the overall redox reaction occurs.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are processes where one substance gains electrons, while another loses them. In a galvanic cell, redox reactions are the driving force for converting chemical energy into electrical energy. Redox reactions compel the movement of electrons, enabling them to do work as they flow.

• Reduction involves gaining electrons, reducing the oxidation state of an element.
• Oxidation, on the other hand, involves the loss of electrons, increasing the oxidation state.
• The overall reaction is a combination of the two half-reactions: one oxidation and one reduction.

Each redox reaction is balanced individually and then combined to derive the complete cell reaction. Understanding redox reactions deepens insight into the operation of galvanic cells and their applications.