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 \(1.0 \mathrm{~atm}\). 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. \(\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)\)

Short answer

a. In the first galvanic cell, the anode is Cr|Cr^3+, and the cathode is Cl2|Cl^-. The electron flow is from the anode (Cr^3+ to Cr_2O_7^2-) to the cathode (Cl2 to Cl^-). The overall balanced equation is \(2Cr^{3+} + 7H_2O + 3Cl_2 \longrightarrow Cr_2O_7^{2-} + 14H^+ + 6Cl^-\). b. In the second galvanic cell, the anode is Mg|Mg^2+, and the cathode is Cu^2+|Cu. The electron flow is from the anode (Mg to Mg^2+) to the cathode (Cu^2+ to Cu). The overall balanced equation is \(Cu^{2+} + Mg \longrightarrow Mg^{2+} + Cu\).

Step-by-step solution

For each overall reaction, we need to identify the redox pair for the two half-cells. A redox pair consists of two species that can be interconverted by the transfer of electrons, which will be the basis for our galvanic cell. a. For the first reaction: 1. \(\text{Redox pair 1: } \mathrm{Cr}^{3+}(a q) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)\) 2. \(\text{Redox pair 2: } \mathrm{Cl}_{2}(g) \rightleftharpoons\mathrm{Cl}^{-}(a q)\) b. For the second reaction: 1. \(\text{Redox pair 1: } \mathrm{Cu}^{2+}(a q) \rightleftharpoons \mathrm{Cu}(s)\) 2. $\text{Redox pair 2: } \mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)$ #Step 2: Identify the anode and cathode for each reaction#

For each reaction, we need to identify which electrode is the anode (where oxidation occurs) and which is the cathode (where reduction occurs). a. For the first reaction: 1. \(\text{Anode (Oxidation): } \mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)\) (1 \(\text{electron}\)) 2. \(\text{Cathode (Reduction): } \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}^{-}(a q)\) (1 \(\text{electron}\)) b. For the second reaction: 1. \(\text{Anode (Oxidation): } \mathrm{Mg}(s) \longrightarrow \mathrm{Mg}^{2+}(a q)\) (2 \(\text{electrons}\)) 2. \(\text{Cathode (Reduction): } \mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Cu}(s)\) (2 \(\text{electrons}\)) #Step 3: Determine the direction of electron flow and balance the overall equation#

The electrons move between the two half-cells, from the anode to cathode. This flow helps to balance the overall equation: a. \(\text{Electron flow: Anode } (\mathrm{Cr}^{3+} \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}) \text{ to the Cathode } (\mathrm{Cl}_{2} \longrightarrow \mathrm{Cl}^{-})\) Overall balanced equation: \(\mathrm{2Cr^{3+}} + \mathrm{7H_2O} + \mathrm{3Cl_2} \longrightarrow \mathrm{Cr_2O_7^{2-}} + \mathrm{14H^+} + \mathrm{6Cl^-}\) b. \(\text{Electron flow: Anode } (\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}) \text{ to the Cathode } (\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu})\) Overall balanced equation: \(\mathrm{Cu^{2+}} + \mathrm{Mg} \longrightarrow \mathrm{Mg^{2+}} + \mathrm{Cu}\) #Step 4: Sketch the galvanic cells #

For each galvanic cell, we need to show the anode and cathode, and the direction of electron flow. a. Galvanic cell sketch: 1. Anode: Cr|Cr^3+ 2. Cathode: Cl2|Cl^- 3. Salt bridge connecting the half-cells 4. External wire connecting the electrodes, with the direction of the electron flow from the anode (Cr^3+ to Cr_2O_7^2-) to the cathode (Cl2 to Cl^-) b. Galvanic cell sketch: 1. Anode: Mg|Mg^2+ 2. Cathode: Cu^2+|Cu 3. Salt bridge connecting the half-cells 4. External wire connecting the electrodes, with the direction of the electron flow from the anode (Mg to Mg^2+) to the cathode (Cu^2+ to Cu)

Key concepts

Understanding Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are fundamental chemical processes where electrons are transferred between atoms or molecules. This electron transfer changes the oxidation states of the reactants. In a redox process, one substance is oxidized (loses electrons) and the other is reduced (gains electrons).

In the context of galvanic cells, we're looking at spontaneous redox reactions, which generate electrical energy. Here's a simplified explanation: As electrons are transferred from the substance being oxidized to the substance being reduced, we can harness this flow of electrons as electric current.

For example, in one of the given reactions, {\(\text{Mg}(s)\)} is oxidized to {\(\text{Mg}^{2+}(aq)\)} as it loses two electrons, while {\(\text{Cu}^{2+}(aq)\)} is reduced to {\(\text{Cu}(s)\)} as it gains those same electrons. It's these types of reactions that power the galvanic cells, providing a venue for learning about the transfer and utilization of energy in chemistry.

Anode and Cathode Identification

Identifying the anode and cathode in a galvanic cell is crucial as they are the sites where oxidation and reduction occur, respectively. The anode is always where oxidation takes place, which means it's where the species loses electrons. Conversely, the cathode is where reduction happens, and the species gains electrons.

In the given examples, for the reaction involving {\(\text{Cr}^{3+}(aq)\)} and {\(\text{Cl}_2(g)\)}, the {\(\text{Cr}^{3+}(aq)\)} is oxidized at the anode, and the {\(\text{Cl}_2(g)\)} is reduced at the cathode. For the {\(\text{Cu}^{2+}(aq)\)} and {\(\text{Mg}(s)\)} reaction, {\(\text{Mg}(s)\)} gives up electrons at the anode, while {\(\text{Cu}^{2+}(aq)\)} accepts electrons at the cathode. These locations are not merely physical spaces but are defined by the chemical processes occurring within the cell.

Direction of Electron Flow in Galvanic Cells

In a galvanic cell, electrons always flow from the anode to the cathode through an external circuit. This movement is due to the potential difference between the two electrodes, driven by the cell's desire to reach equilibrium. The electron flow's direction is crucial for powering electronic devices or anything connected to the cell.

In our example reactions, for the {\(\text{Cr}^{3+}(aq)\)} and {\(\text{Cl}_2(g)\)} reaction, electrons flow from the anode (where {\(\text{Cr}^{3+}\)} is oxidized to {\(\text{Cr}_2\text{O}_7^{2-}\)}) to the cathode (where {\(\text{Cl}_2\)} is reduced to {\(\text{Cl}^-\)}). Similarly, for the {\(\text{Cu}^{2+}(aq)\)} and {\(\text{Mg}(s)\)} reaction, the electrons travel from magnesium, being oxidized at the anode, to copper, being reduced at the cathode. Following the right direction of electron flow in such sketches and understanding its implications are essential for mastering electrochemistry concepts.