University:
Santa Ana College
Course:
CHEM 229 | General Chemistry and Qualitative Analysis
Academic year:

2021
Views:

196

Pages:

2
Author:

Krista Hamilton

Study Guide for Electrochemistry How to find K from E°cell 1. To review: • Cell potentials are intensive properties, meaning that they don’t depend on the amount of material. For example: Pb2+ + 2e- -> Pb(s) E° = -0.125 2Pb2+ + 4e- -> 2Pb(s) E° = -0.125 Standard conditions for electrochemical cells: o Pure solids and liquids o 1 M solutions o 1 bar (~ 1 atm) gases • Le Chatelier’s Principle: Changing the conditions of a reaction at equilibrium will cause the reaction to produce more reactant or product until equilibrium is reestablished. 2. Relating E°cell to K • Standard conditions apply for both E°cell and K • From thermodynamics: • ΔG° = -RTlnK • From previous electrochemistry discussions: ΔG° = -nFE°cell • Putting both together: -RTlnK = -nFE°cell E°cell = (RT/nF)lnK • • Where: o E°cell is the standard cell potential (in V) o R is the thermodynamic form of the gas constant (R = 8. 3145 J*mol-1*K-1) o T is the temperature (in K) o n is the number of moles of electrons that are transferred in the reaction o F is Faraday’s constant (F = 96485 C*mol-1) o K is the equilibrium constant for the reaction Usually, the cell is at room temperature (298 K). This reduces the equation to: E°cell = (0.025693/n)*lnK 3. Example: Using the following standard reduction potentials: Ag+ + e- <-> Ag(s) E° = 0.799V Ag2Mo4(s) + 2e- <-> 2Ag(s) + MoO42- E° = 0.486 V Calculate the solubility product constant for Ag2MoO4(s) Solution: • We know that the solubility product constant equation has this form: Ag2MoO4(s) <-> 2Ag+(aq) + MoO42+(aq) • To get to this equation, we first have to flip one of the half-reactions (changing it from reduction to oxidation) and multiply it by 2. After that we can add the half-reactions. 2Ag(s) <-> 2Ag+ + 2eAg2MoO4(s) + 2e- <-> 2Ag(s) + MoO42--------------------------------------------------Ag2MoO4(s) <-> 2Ag+ + MoO42- We now have the 2 half-reactions. The oxidation half-reaction’s E° becomes E°anode. The reduction half-reaction’s E° becomes E°cathode. We can then find E°cell: E°cell = E°cathode - E°anode E°cell = 0.486 V - 0.799 V E°cell = -0.313 V • We can now find K: E°cell = (0.025693/n)*lnK E°cell *(n/0.025693) = lnK eE°cell *(n/0.025693) = K e-0.313*(2/0.025693) = K K = 2.622*10-11 • Conceptually, this makes sense. We should expect a non-spontaneous reaction (one with a negative E°cell) to produce very little, if any product at equilibrium (to have a very small K)

Electrochemistry Study Guide 3

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