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 cell consists of two half-cells connected by a salt bridge. At the anode (left), oxidation occurs with the reaction Cr -> Cr^3+ + 3e^-, while at the cathode (right), reduction occurs with the reaction Fe^2+ + 2e^- -> Fe. The overall balanced cell equation is 2Cr + 6Fe^2+ -> 2Cr^3+ + 6Fe, and the cell voltage is -1.18 V. Electrons flow from the anode (chromium) to the cathode (iron).

Step-by-step solution

Write the half-reactions for both iron and chromium

We need to convert iron(II) to iron metal and change chromium metal to chromium(III). For iron: Fe^2+ + 2e^- -> Fe (reduction half-reaction) For chromium: Cr -> Cr^3+ + 3e^- (oxidation half-reaction)

Determine the standard reduction potentials

Look up the standard reduction potentials for both half-reactions: Fe^2+ + 2e^- -> Fe (E° = -0.44 V) Cr^3+ + 3e^- -> Cr (E° = -0.74 V) Since we want to convert chromium-metal to chromium(III), we need to reverse the chromium half-reaction: Cr -> Cr^3+ + 3e^- (E° = 0.74 V)

Calculate the cell potential

The cell potential (E°(cell)) can be calculated as: E°(cell) = E°(cathode) - E°(anode) Since the iron half-reaction has a higher reduction potential, it will be the cathode (reduction occurs at the cathode): E°(cell) = (-0.44 V) - (0.74 V) = -1.18 V

Sketch the cell, show electron flow, and label the anode and cathode

- Left: Anode (Chromium): Write the oxidation half-reaction: Cr -> Cr^3+ + 3e^- Label chromium as the anode (A) Electrons flow from anode to cathode - Right: Cathode (Iron): Write the reduction half-reaction: Fe^2+ + 2e^- -> Fe Label iron as the cathode (C) - Connect the two half-cells with a salt bridge. A salt bridge is essential to maintain electrical neutrality in both half-cells by allowing the exchange of ions from one half-cell to the other.

Balance the overall cell equation and calculate the voltage

To balance the overall cell equation, multiply each half-reaction by the appropriate number to equalize the number of electrons exchanged: Oxidation: Cr -> Cr^3+ + 3e^- (x2) Reduction: Fe^2+ + 2e^- -> Fe (x3) Now add them up: 2Cr + 6Fe^2+ -> 2Cr^3+ + 6Fe The voltage of the cell is -1.18 V.