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

In the galvanic cell, the anode half-reaction is : \(3Fe^{2+}(aq) \rightarrow 3Fe(s) + 6e^{-}\) and the cathode half-reaction is: \(2Cr(s) + 6e^{-} \rightarrow 2Cr^{3+}(aq)\). The cell potential (voltage) is -1.18 V. The balanced overall cell equation is: \(3Fe^{2+}(aq) + 2Cr(s) \rightarrow 3Fe(s) + 2Cr^{3+}(aq)\). The electron flow is from anode (iron half-cell) to cathode (chromium half-cell).

Step-by-step solution

Identify Half-Reactions and Electrodes

Anode (where oxidation occurs): Iron(II) (Fe^2+) is oxidized to form iron metal (Fe) $$Fe^{2+}(aq) \rightarrow Fe(s) + 2e^{-}$$ Cathode (where reduction occurs): Chromium metal (Cr) changes to chromium(III) (Cr^3+) $$Cr(s) + 3e^{-} \rightarrow Cr^{3+}(aq)$$

Calculate the Cell Potential

We can find the cell potential by using the standard reduction potentials of the two half-reactions. Standard reduction potential for Fe^2+ to Fe: -0.44 V Standard reduction potential for Cr^3+ to Cr: -0.74 V $$E_{cell} = E_{cathode} - E_{anode}$$ When we consider the oxidation and reduction process in the cell, reverse the anode reaction first: $$Fe(s) - 2e^{-} \rightarrow Fe^{2+}(aq)$$ Now, the standard reduction potential for the anode becomes +0.44 V (reversal). $$E_{cell} = (-0.74) - (+0.44) = -1.18 V$$

Balance the Overall Cell Equation

In order to balance the overall cell equation, we need to balance both charge and atoms. We will balance the electrons first. For the anode half: $$Fe^{2+}(aq) \rightarrow Fe(s) + 2e^{-}$$ For the cathode half: $$Cr(s) + 3e^{-} \rightarrow Cr^{3+}(aq)$$ Multiply the first half-reaction by 3 and second half-reaction by 2 to get the same number of electrons exchanged in both half-reactions. Anode half-reaction (x3): $$3Fe^{2+}(aq) \rightarrow 3Fe(s) + 6e^{-}$$ Cathode half-reaction (x2): $$2Cr(s) + 6e^{-} \rightarrow 2Cr^{3+}(aq)$$ Now, add these balanced half-reactions to get the overall cell equation: $$3Fe^{2+}(aq) + 2Cr(s) \rightarrow 3Fe(s) + 2Cr^{3+}(aq)$$

Sketch the Cell and Label Components

In the sketch: - Label the anode side: "Anode: Fe^2+ to Fe". - Label the cathode side: "Cathode: Cr to Cr^3+". - Label the direction of electron flow from anode to cathode (electrons are produced at the anode and consumed at the cathode). - Include a salt bridge which allows the movement of ions to keep the cell electrically neutral (e.g., Na^+ and SO_4^2- ions). - Label the cell potential as "E = -1.18 V". The sketched cell should clearly show the electron flow, anode, cathode, and each half-reaction with their respective labels.