For the reaction that occurs in the voltaic cell,
Fe(s)|Fe2+(aq)||Cr3+(aq), Cr2+(aq)|Pt(s)
determine E°cell.
I don't understand how the Pt(s) would be used to solve this question.
Please help!
Thanks for explaining what you don't understand. It makes it easier. The Pt is an inert electrode. Obviously you can't make electrodes of Cr^+3 and/or Cr^+2 (as you can with iron meta); therefore, an inert electrode is used for electrical contact. Otherwise, the Nernst equation is used to calculate Eo of the cell.
To determine E°cell, you can use the standard reduction potentials (E°red) of the half-reactions involved. The Pt(s) in the voltaic cell is used as an inert electrode, which means it does not participate in the reaction itself but serves as a conductive surface to facilitate electron transfer.
In the given voltaic cell diagram, the half-reactions are as follows:
1. Cathode (reduction half-reaction): Cr3+(aq) + 3e- → Cr(s)
2. Anode (oxidation half-reaction): Fe(s) → Fe2+(aq) + 2e-
To determine the overall cell potential (E°cell), you need to calculate the difference between the reduction potentials of the cathode and anode. The reduction half-reaction with the more positive E°red will be the cathode, and the other half-reaction will be the anode.
You can find the standard reduction potentials (E°red) in a table of standard reduction potentials. Make sure the values are for the same conditions as your reaction (e.g., temperature, concentration). Let's assume the values are:
E°red for Cr3+ + 3e- → Cr: -0.74 V
E°red for Fe2+ + 2e- → Fe: -0.44 V
Now, subtract the E°red for the anode from the E°red for the cathode to find E°cell:
E°cell = E°red(cathode) - E°red(anode)
= -0.74 V - (-0.44 V)
= -0.3 V
Therefore, the E°cell for the given voltaic cell is -0.3 V.