Problem 7 - Chemistry: Principles and Reactions

Question

Consider a salt bridge cell in which the anode is a manganese rod immersed in an aqueous solution of manganese(II) sulfate. The cathode is a chromium strip immersed in an aqueous solution of chromium(III) sulfate. Sketch a diagram of the cell, indicating the flow of the current throughout. Write the half- equations for the electrode reactions, the overall equation, and the abbreviated notation for the cell.

Step-by-Step Solution

Answer

Answer: The half-reactions are: Anode: Mn(s) -> Mn^{2+}(aq) + 2e^- Cathode: Cr^{3+}(aq) + 3e^- -> Cr(s) The overall equation is: 3Mn(s) + 2Cr^{3+}(aq) -> 3Mn^{2+}(aq) + 2Cr(s) The abbreviated notation for the cell is: Mn(s)|Mn^{2+}(aq)||Cr^{3+}(aq)|Cr(s)

Step 1: Sketch a Diagram of the Cell

Draw a diagram with a manganese (Mn) rod as the anode and a chromium (Cr) strip as the cathode. The anode should be immersed in an aqueous solution of manganese(II) sulfate (MnSO4), and the cathode should be immersed in an aqueous solution of chromium(III) sulfate (Cr2(SO4)3). Connect the anode and the cathode with a wire, and also connect them with a salt bridge, which allows the flow of ions but not electrons.

Step 2: Identify the Redox Reactions

In this cell, the anode (Mn rod) undergoes oxidation, while the cathode (Cr strip) undergoes reduction. The oxidation half-reaction occurring at the anode is: Mn(s) -> Mn^{2+}(aq) + 2e^- The reduction half-reaction occurring at the cathode is: Cr^{3+}(aq) + 3e^- -> Cr(s)

Step 3: Write the Overall Equation

To combine the two half-equations and get the overall equation, we must ensure that the number of electrons in both half-reactions is the same. To achieve this, multiply the first half-reaction by 3 and the second half-reaction by 2: 3Mn(s) -> 3Mn^{2+}(aq) + 6e^- 2Cr^{3+}(aq) + 6e^- -> 2Cr(s) Now we can add both half-reactions: 3Mn(s) + 2Cr^{3+}(aq) -> 3Mn^{2+}(aq) + 2Cr(s)

Step 4: Write the Abbreviated Notation for the Cell

In the abbreviated notation, the anode is written first, followed by its solution, then the cathode and its solution. For the anode, the half-reaction is: Mn(s) -> Mn^{2+}(aq) For the cathode, the half-reaction is: Cr^{3+}(aq) -> Cr(s) So, the abbreviated notation is: Mn(s)|Mn^{2+}(aq)||Cr^{3+}(aq)|Cr(s) In summary, the cell consists of a manganese anode and a chromium cathode connected by a salt bridge, with the following half-reactions: Anode: Mn(s) -> Mn^{2+}(aq) + 2e^- Cathode: Cr^{3+}(aq) + 3e^- -> Cr(s) The overall equation is: 3Mn(s) + 2Cr^{3+}(aq) -> 3Mn^{2+}(aq) + 2Cr(s) And the abbreviated notation for the cell is: Mn(s)|Mn^{2+}(aq)||Cr^{3+}(aq)|Cr(s)

Key Concepts

Redox Reactions

In electrochemical cells, redox reactions are fundamental processes. A redox reaction involves the transfer of electrons between two species. One species undergoes oxidation, which means it loses electrons, while the other undergoes reduction, gaining electrons. In our example, the manganese (Mn) rod serves as the anode, where it oxidizes, releasing electrons and transforming into Mn\(^{2+}\). Meanwhile, the chromium at the cathode receives these electrons, reducing Cr\(^{3+}\) to form solid chromium (Cr). This simultaneous process ensures the flow of electrons, generating an electric current.

• Oxidation occurs at the anode: Mn(s) \(\rightarrow\) Mn\(^{2+}\)(aq) + 2e\(^{-}\)
• Reduction takes place at the cathode: Cr\(^{3+}\)(aq) + 3e\(^{-}\) \(\rightarrow\) Cr(s)

These reactions power many types of batteries and are crucial in metal corrosion and biochemical processes.

Half-Equations

Half-equations are simplified representations of redox reactions showing only oxidation or reduction processes. Each half-equation isolates one part of the complete redox process, either loss or gain of electrons. They are useful for understanding the behavior of substances at each electrode, predicting overall cell reactions, and balancing redox reactions.
In our manganese and chromium cell:

• Oxidation at anode: Mn(s) \(\rightarrow\) Mn\(^{2+}\)(aq) + 2e\(^{-}\)
• Reduction at cathode: Cr\(^{3+}\)(aq) + 3e\(^{-}\) \(\rightarrow\) Cr(s)

To derive the overall cell equation, match the number of electrons from the half-reactions by scaling them and then sum them up. Here, multiplying the manganese reaction by 3 and the chromium reaction by 2 does the trick, providing a balanced reaction:
3Mn(s) + 2Cr\(^{3+}\)(aq) \(\rightarrow\) 3Mn\(^{2+}\)(aq) + 2Cr(s).

Salt Bridge

The salt bridge is a key component in an electrochemical cell. It allows ions to move between the two half-cells, maintaining electrical neutrality. Without it, the flow of electrons would quickly halt as charge builds up at each electrode.
A salt bridge usually contains a gel saturated with a salt solution, such as KCl or Na\(^{+}\)NO\(^{-}\). The ions can flow to complete the electrical circuit, balancing out the charges that accumulate when oxidation and reduction occur.
In our cell:

• Anions, affiliated with the reduction reaction at the cathode, migrate to the anode.
• Cations, associated with oxidation at the anode, move towards the cathode.

These ion movements via the salt bridge ensure a steady current flow and prolong the cell's operation.

Cell Notation

Cell notation is a shorthand method to describe a galvanic cell, simplifying its electron transfer process. It provides a quick glance at the components involved in the electrochemical reaction without detailed diagrams.
The standard format includes writing the anode on the left, separated by a double line from the cathode on the right. Each half-cell is divided by a single line, indicating a phase boundary or electrode-solution interface.
In our example:

• Anode: Mn(s), Anode solution: Mn\(^{2+}\)(aq)
• Cathode solution: Cr\(^{3+}\)(aq), Cathode: Cr(s)

This yields the notation:
\[ \text{Mn(s) | Mn}^{2+}\text{(aq) || Cr}^{3+}\text{(aq) | Cr(s)} \]The double vertical line represents the salt bridge or junction between the two half-cells.