Analytical Chemistry Cell Potential

Question

A beaker contains a solution of 0.0200 M MnO4 – and 0.00500 M Mn2+ at a pH of 0.0. A second beaker contains 0.150 M Fe2+ and 0.00150 M Fe3+. A salt bridge connects the two beakers, and Pt electrodes are placed in each. The black (-) lead is hooked up to the Pt electrode associated with the MnO4 – / Mn2+ beaker. What would be the measured cell potential at the start of the reaction and after the reaction reaches equilibrium?

MnO4 ¯ + 8 H+ + 5e¯ ⇋ Mn2+ +4H2O Eo = 1.507 V

Fe3+ + e¯ ⇋ Fe2+ Eo = 0.771 V

How do I solve questions like this? Is there a method to follow to help me understand this better? I'm struggling to get questions like this correct and I'm kind of all over the place trying to solve them.

Answer 1

To solve for the cell potential at the start of the reaction, we must first determine the overall, balanced reaction that occurs in this galvanic cell:

The half reactions are (1) MnO₄^- (aq) + 8 H⁺(aq) + 5e^- ↔ Mn²⁺ (aq) + 4 H₂O (l) E^o= 1.507V

(2) Fe²⁺ (aq) ↔ Fe³⁺ (aq) + e^- E^o= -0.771V

Notice that reaction 2 is written as an oxidation half-reaction. This is because MnO₄^- is a stronger oxidizing agent than Fe³⁺, as seen by their relative standard reduction potentials.

The overall reaction is then MnO₄^- (aq) + 8 H⁺(aq) + 5Fe²⁺ (aq) ↔ Mn²⁺ (aq) + 4 H₂O (l) + 5Fe³⁺ (aq)

which has an E^o_cell = E^o_cathode - E^o_anode = 1.507V - 0.771 V = 0.736V. As a note, the oxidation half-reaction has been multiplied by five in order to balance the electrons.

Now that we have our overall reaction, we can calculate the cell potential using the Nernst-equation:

E_cell = E^o_cell - (0.0591/n)log(Q) at 25^oC where n is the # of electrons transferred per reaction and Q is the reaction quotient.

E_cell = 0.736V - (0.0591/5)log(([Mn²⁺][Fe³⁺]⁵)/([MnO₄^-][H⁺]⁸[Fe²⁺]⁵))

E_cell = 0.736V - (0.0591/5)log(((.005)(.0015)⁵)/((.02)(1)⁸(.15)⁵))

E_cell = 0.736V - (0.0591/5)log(2.5x10^-11)

E_cell = 0.736V + 0.125V = 0.861V

This is the cell potential before any current flow has occurred. When the galvanic cell is allowed to discharge, reactants will be consumed and products will be produced until equilibirum is reached. At equilibrium, there is no more current flow and E_{cell equilibrium} = 0.

The formula ΔG_rxn = -nFE_cell can be used to help understand why this is the case. At the start of the reaction, the free energy difference between the products and reactants in the cell is at its largest. This corresponds to a large, positive E_cell. As the reaction proceeds to equilibrium, ΔG_rxn decreases in magnitude as reactants are converted into products until it becomes zero (when ΔG_rxn = 0, E_cell = 0). At equilibrium, there is no chemical driving force to push electrons through the wire and the cell is considered "dead."

The steps I used in this problem can be utilized when solving other electrochemistry problems. My best advice, however, is to obtain a thorough understanding of the concepts at play.