Electrochemistry Study Guide 4 | Study Guide

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

2021
Views:

281

Pages:

2
Author:

Giada Caldwell

Study Guide for Electrochemistry How to find cell potentials at non-standard conditions: 1. Things to remember: • From thermodynamics: ΔG = ΔG° + RTlnQ • • Where: o ΔG is the Gibbs free energy (in units of J*mol-1) o ΔG° is the Gibbs free energy at standard conditions (in units of J*mol-1) o R is the thermodynamic form of the gas constant (R = 8.3145 J*mol-1*K-1) o T is the temperature (in units of K) o Q is the reaction quotient (unitless) From previous electrochemistry discussions: ΔG° = -nFE°cell • Similarly: ΔG = -nFEcell • Where: o n is the number of moles of electrons transferred in the redox reaction o F is Faraday’s constant (F = 96385 C*mol-1) o E°cell is the cell potential at standard conditions o Ecell is the cell potential at non-standard conditions (what we want) 2. The derivation: • Putting things together: -nFEcell = -nFE°cell +RTlnQ -nFEcell/-nF = -nFE°cell/-nF +RTlnQ/-nF Ecell = E°cell - RT/nF * lnQ • Since reactions usually take place at 25°C (298 K), this can usually be reduced to: Ecell = E°cell - 0.025693/n * lnQ • You might also see this equation as the Nernst equation: Ecell = E°cell - 0.0592/n * logQ 3. Example: Determine the value of Ecell for the following cell at room temperature: Pt|Fe2+(aq) (0.10 M), Fe3+(aq) (0.20 M)|| Ag+(aq) (1.0 M)|Ag(s) 1. Rewrite the cell as a set of redox half-reactions: Cathode (Red.): Ag+(aq) + e- -> Ag(s) Anode (Ox.): Fe2+(aq) -> Fe3+(aq) + e2. Find the standard reduction potential for each half-reaction (from a table): Cathode (Red.): Ag+(aq) + e- -> Ag(s) E°cathode = 0.800 V Anode (Ox.): Fe2+(aq) -> Fe3+(aq) + e- E°anode = 0.771 V 3. Find E°cell: E°cell = E°cathode - E°anode E°cell = 0.800 V – 0.771 V E°cell = 0.029 V 4. Find n: Because in this reaction 1 electron is gained for every 1 electron that is lost, the total electrons transferred is 1 mole. 5. Find Q: • Overall reaction: Ag+(aq) + Fe2+(aq) ->Ag(s) + Fe3+(aq) • General Q equation: Q = [Fe3+]/[Ag+][Fe2+] • Plugging in concentrations: Q = (0.20)/((1.0)(0.10) Q=2 6. Find Ecell: Ecell = E°cell - 0.0592/n * logQ Ecell = E°cell - 0.0592/1 * log(2) Ecell = 0.029 - 0.0592 * 0.30103 Ecell = 0.011 V

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