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General Chemistry Principles, Patterns, and Applications, 2011

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Solving the last expression for ΔG° for the overall half-reaction,<br />

Equation 19.56<br />

ΔG° = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V)<br />

Three electrons (n = 3) are transferred in the overall reaction (Equation 19.55), so substituting<br />

into Equation 19.52 <strong>and</strong> solving for E° gives the following:<br />

DG°F(0.13 V)E° = -nFE cell =<br />

-(3)(F)(E cell) = -0.13 V 3 = -0.043 V<br />

This value of E° is very different from the value that is obtained by simply adding the potentials for the<br />

two half-reactions (0.32 V) <strong>and</strong> even has the opposite sign.<br />

Note the Pattern<br />

Values of E° for half-reactions cannot be added to give E° for the sum of the half-reactions; only values of<br />

ΔG° = −nFE°cell for half-reactions can be added.<br />

The Relationship between Cell Potential <strong>and</strong> the Equilibrium Constant<br />

We can use the relationship between ΔG° <strong>and</strong> the equilibrium constant K, defined inChapter 18 "Chemical<br />

Thermodynamics", to obtain a relationship between E°cell <strong>and</strong> K. Recall that for a general reaction of the<br />

type aA + bB → cC + dD, the st<strong>and</strong>ard free-energy change <strong>and</strong> the equilibrium constant are related by the<br />

following equation:<br />

Equation 19.57<br />

ΔG° = −RT ln K<br />

Given the relationship between the st<strong>and</strong>ard free-energy change <strong>and</strong> the st<strong>and</strong>ard cell potential (Equation<br />

19.52), we can write<br />

Equation 19.58<br />

−nFE° cell = −RT ln K<br />

Rearranging this equation,<br />

Equation 19.59<br />

E cell = (RTnF)ln K<br />

For T = 298 K, Equation 19.59 can be simplified as follows:<br />

Equation 19.60<br />

E cell = (RTnF)ln K = [8.314 J / (mol ×K)<br />

(298K)]n[96,486 J / (V × mol)]2.303 log K = (0.0591 Vn)log K<br />

Saylor URL: http://www.saylor.org/books<br />

Saylor.org<br />

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