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. \(C r^{3+}(a q)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)\)

Short answer

For reaction (a): The galvanic cell consists of Cr^3+ ions being reduced to Cr2O7^2- ions at the cathode, and Cl2 gas being oxidized to Cl^- ions at the anode. The electrons flow from the anode to the cathode. The overall balanced equation is: \(2 \: Cr^{3+} + 7 \: H2O + 3 \: Cl2 \rightarrow Cr2O7^{2-} + 14 \: H^+ + 6 \: Cl^-\). For reaction (b): The galvanic cell consists of Cu^2+ ions being reduced to Cu solid at the cathode, and Mg solid being oxidized to Mg^2+ ions at the anode. The electrons flow from the anode to the cathode. The overall balanced equation is given: \(Cu^{2+} + Mg \rightarrow Mg^{2+} + Cu\).

Step-by-step solution

Break down the reaction into the half-reactions

We need to identify which component is being reduced and which is being oxidized. From the overall reaction, we can deduce that: - Chromium ions (Cr^3+) are being reduced to form Chromium ion-dichromate (Cr2O7^2-) - Chlorine gas (Cl2) is being oxidized to form Chloride ions (Cl^-) So, we can write the half-reactions: - Reduction: \( 2 \: Cr^{3+} + 7 \: H2O \rightarrow Cr2O7^{2-} + 14 \: H^+ + 6 \: e^-\) - Oxidation: \( Cl2 + 2 \: e^- \rightarrow 2 \: Cl^- \) Now, we can find the balanced overall equation.

Balance the overall equation

Multiply the oxidation reaction by 3 to make sure the number of electrons lost in oxidation equals the number gained in the reduction: - Reduction: \( 2 \: Cr^{3+} + 7 \: H2O \rightarrow Cr2O7^{2-} + 14 \: H^+ + 6 \: e^-\) - Oxidation (multiplied by 3): \( 3 \: Cl2 + 6 \: e^- \rightarrow 6 \: Cl^- \) Now we can add the two half-reactions to obtain the overall balanced equation: \[2 \: Cr^{3+} + 7 \: H2O + 3 \: Cl2 \rightarrow Cr2O7^{2-} + 14 \: H^+ + 6 \: Cl^-\]

Sketch the galvanic cell

Now that we have the half-reactions and overall equation, we can sketch the galvanic cell. - The cathode is where reduction occurs, so Cr^3+ ions are reduced at the cathode. - The anode is where oxidation occurs, so Cl2 gas is oxidized at the anode. - The electrons flow from the anode (oxidation) to the cathode (reduction). #For reaction (b): Cu^2+(aq)+Mg(s)↔Mg^2+(aq)+Cu(s)#

Break down the reaction into the half-reactions

We can deduce that: - Magnesium solid (Mg) is being oxidized to form Magnesium ions (Mg^2+) - Copper ions (Cu^2+) are being reduced to form Copper solid (Cu) So, we can write the half-reactions: - Reduction: \( Cu^{2+} + 2 \: e^- \rightarrow Cu \) - Oxidation: \( Mg \rightarrow Mg^{2+} + 2 \: e^- \) Here, the overall balanced equation is given.

Sketch the galvanic cell

Now that we have the half-reactions and overall equation, we can sketch the galvanic cell. - The cathode is where reduction occurs, so Cu^2+ ions are reduced at the cathode. - The anode is where oxidation occurs, so Mg solid is oxidized at the anode. - The electrons flow from the anode (oxidation) to the cathode (reduction).

Key concepts

Oxidation-Reduction Reactions

In a galvanic cell, the magic happens through oxidation-reduction reactions, also known as redox reactions. These comprise two interlinked processes: oxidation (loss of electrons) and reduction (gain of electrons). Think of it as a transaction: one element donates electrons (oxidation), while another accepts them (reduction).

For example, in reaction (a), we see chromium ions ( ext{Cr}^{3+}) being reduced while chlorine gas ( ext{Cl}_2) gets oxidized. This exchange drives the current in a galvanic cell. Similarly, for reaction (b), magnesium ( ext{Mg}) undergoes oxidation as it loses electrons, while copper ions ( ext{Cu}^{2+}) undergo reduction as they gain electrons.

Understanding these electron transfers is crucial to grasp how energy conversion occurs in these cells, making redox reactions the backbone of chemical power in galvanic cells.

Half-Reactions

Half-reactions are a handy way to break down the overall process of a redox reaction into manageable pieces. Each half-reaction shows either the oxidation or reduction process specifically.

Let's consider reaction (a) again: the reduction half-reaction is \(2 \: ext{Cr}^{3+} + 7 \, ext{H}_2 ext{O} \rightarrow ext{Cr}_2 ext{O}_7^{2-} + 14 \, ext{H}^+ + 6 \, ext{e}^-\), and the oxidation half-reaction is \(3 \, ext{Cl}_2 + 6 \, ext{e}^- \rightarrow 6 \, ext{Cl}^-\). Each separately balances the electrons gained or lost.

These half-reactions provide a clearer depiction of where the losses and gains occur, making it simple to see which compound functions as the reducing agent and which as the oxidizing agent.

Electrode Identification

In any galvanic cell, electrodes are crucial for the reaction's progress. The oxidizing and reducing agents interact at these conductive terminals.

Each galvanic cell has two electrodes: the anode and the cathode. The anode is where oxidation occurs, meaning it's our electron donor. Conversely, the cathode is where reduction takes place, acting as our electron acceptor. For example, in reaction (b), magnesium ( ext{Mg}) acts as the anode, whereas copper ( ext{Cu}) serves as the cathode.

Remember: "An Ox" stands for Anode Oxidation, and "Red Cat" for Reduction Cathode. This simple mnemonic can help you keep track of the electron flow in an electrochemical cell.

Electron Flow

Electron flow in a galvanic cell is fundamental as it represents the movement of charge, enabling electrical energy generation. Electrons naturally move from the anode to the cathode due to the potential difference between the two electrodes.

In the reactions discussed, electrons flow from where oxidation releases them at the anode – magnesium in reaction (b) and chlorine gas in reaction (a) – and travel towards the cathode, where reduction welcomes them (copper in reaction (b) and chromium ions in reaction (a)).

This unidirectional flow of electrons from the anode to the cathode through an external circuit is what powers devices connected to the galvanic cell, harnessing the chemical reaction into usable electrical energy.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reaction where reactants and products are formed at the same rate, leading to no net change in concentrations over time.

In the context of galvanic cells and redox reactions, equilibrium is significant as it dictates how efficiently a cell can perform. An equilibrium shift can impact the cell's voltage and functionality.

For example, changes in conditions like concentration or pressure can shift equilibrium, potentially leading to less efficient electron flow and reduced power output. In galvanic cells, Le Chatelier's principle elucidates these shifts, as the system adjusts to counterbalance deviations from equilibrium, maintaining optimal energy conversion conditions.