Question

Sketch a cell that forms iron metal from iron(II) while changing chromium metal to chromium(III). Calculate the voltage, show the electron flow, label the anode and cathode, and balance the overall cell equation.

Short answer

The electrochemical cell consists of an anode (chromium) where oxidation occurs and a cathode (iron(II)) where reduction occurs. The balanced cell equation is \(2Cr + 3Fe^{2+} \rightarrow 2Cr^{3+} + 3Fe\), and the cell potential is 0.30 V. Electrons flow from chromium (anode) to iron (cathode). The cell can be sketched with two compartments connected by a salt bridge, and the electron flow is represented with arrows from the anode to the cathode.

Step-by-step solution

Identify Half Reactions

To start, we need to identify the half reactions involved in this electrochemical cell. The conversion of iron(II) to iron metal is a reduction half-reaction, while the conversion of chromium metal to chromium(III) is an oxidation half-reaction. The half reactions are: Reduction: \(Fe^{2+} + 2e^- \rightarrow Fe \) Oxidation: \(Cr \rightarrow Cr^{3+} + 3e^-\)

Balance the Half Reactions

In order to find the overall balanced cell equation, we need to balance the half reactions. We need to balance the number of electrons in both half reactions. Let's multiply the reduction half-reaction with 3, and the oxidation half-reaction with 2, which will make the total number of electrons equal in both equations: Reduction: \(3(Fe^{2+} + 2e^- \rightarrow Fe) \Rightarrow 3Fe^{2+} + 6e^- \rightarrow 3Fe\) Oxidation: \(2(Cr \rightarrow Cr^{3+} + 3e^-) \Rightarrow 2Cr \rightarrow 2Cr^{3+} + 6e^-\) Now we can combine these two balanced half-reactions to form the overall balanced cell equation.

Overall Balanced Cell Equation

Adding the balanced half-reactions together, we get the overall balanced cell equation: \(2Cr + 3Fe^{2+} \rightarrow 2Cr^{3+} + 3Fe\)

Determine Cell Potential

Now we need to determine the cell potential (voltage) of the cell. This can be done by finding both the reduction and oxidation potentials and applying the formula: \(E_{cell} = E_{cathode} - E_{anode}\) From a standard reduction potential table, we can find the values for the two half-reactions: Reduction potential for \(Fe^{2+}: E^0_{red,Fe} = -0.44 V\) Oxidation potential for \(Cr: E^0_{ox,Cr} = -0.74 V\) So the cell potential is: \(E_{cell} = (-0.44) - (-0.74) = 0.30 V\)

Identify Anode, Cathode, and Electron Flow

In an electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode. Therefore: Anode: Chromium (oxidation) \(Cr \rightarrow Cr^{3+} + 3e^-\) Cathode: Iron(II) (reduction) \(Fe^{2+} + 2e^- \rightarrow Fe\) Electron flow: The electrons flow from the anode (chromium) to the cathode (iron) for the cell to function.

Sketch and Label the Cell

To sketch and label the cell, follow these steps: 1. Draw two separate compartments representing the anode and cathode half-cells. 2. In the anode compartment (left), place the chromium electrode and label it as "Anode" and write the oxidation half-reaction. 3. In the cathode compartment (right), place the iron electrode and label it as "Cathode" and write the reduction half-reaction. 4. Connect the two compartments with a salt bridge to maintain electrical neutrality. 5. On the diagram, indicate the electron flow with arrows from the anode (Chromium) to the cathode (Iron). Now you have sketched the electrochemical cell for the formation of iron metal from iron(II) while changing chromium metal to chromium(III), calculated the cell voltage, and labeled the anode and cathode.

Key concepts

Oxidation Half-Reaction

The oxidation half-reaction is a crucial concept in electrochemistry, representing the loss of electrons by a species. In the step by step solution provided, chromium serves as an example, undergoing oxidation from its metallic state to a chromium ion with a charge of +3. The reaction can be represented as:
\( Cr \rightarrow Cr^{3+} + 3e^- \).
In this process, chromium loses three electrons and thereby gets oxidized. To ensure students grasp this concept, it's important to highlight that oxidation always involves an increase in oxidation state, which is associated with the element becoming more positively charged due to the loss of negatively charged electrons.

Reduction Half-Reaction

The reduction half-reaction occurs simultaneously with the oxidation process in an electrochemical cell. It involves the gain of electrons, leading to a decrease in oxidation state. Taking iron(II) as an example in our given solution, the reduction half-reaction is depicted as:
\( Fe^{2+} + 2e^- \rightarrow Fe \).
The iron(II) ions are reduced as they gain two electrons to form solid iron. A helpful tip for students is to remember that reduction is the opposite of oxidation—often remembered by the mnemonic 'LEO goes GER', where LEO stands for 'Loss of Electrons is Oxidation' and GER means 'Gain of Electrons is Reduction'. This helps clarify that in a reduction half-reaction, a substance's oxidation state decreases.

Cell Potential Calculation

The cell potential, or voltage, of an electrochemical cell, provides a quantitative measure of the cell's ability to produce an electric current. The cell's voltage is calculated using the half-reaction potentials:
\( E_{cell} = E_{cathode} - E_{anode} \).
The reduction half-reaction at the cathode has a potential of -0.44V for iron, while the oxidation potential for chromium at the anode is -0.74V. Thus, cell potential is calculated as:
\( E_{cell} = (-0.44) - (-0.74) = 0.30 V \).
This positive voltage indicates a spontaneous reaction. It's pivotal for students to understand that the larger the positive value of cell potential, the more spontaneous the reaction.

Anode and Cathode Identification

Identifying the anode and cathode is key to understanding the flow of electrons in an electrochemical cell. The anode is always where oxidation occurs—where electrons are lost. In our example, chromium is oxidized, making it the anode. The cathode is the site of reduction, where electrons are gained; this is where iron(II) is reduced to iron metal.
In a cell diagram, the anode is typically placed on the left and the cathode on the right with the salt bridge connecting them. The external circuit reflects the physical path of electron flow which occurs from the anode to the cathode. Emphasizing the roles of each component and the direction of electron flow is instrumental for students to visualize and understand the function of the entire electrochemical cell.