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Redox reactions like all others can reach an equilibrium state. Since we have a relationship between E0 and DG0 as well as one between DG0 and K, we can derive a relationship between the cell voltage and the equilibrium constant. Since we have

  • DG0 = -nFE0
  • DG0 = -RTln(K)
we can combine the two equations into one:
E0 = (RT/nF)*ln(K)

At 25oC, the term RT/F has the value of 0.0257 V, so we can simplify the above equation at 25oC

E0 = (0.0257/n)*ln(K)
With the above equations, we can derive the value of the cell voltage from the equilibrium constant and vice versa

Example: The copper-zinc redox reaction shown below has an E0 value of 1.101 V. What is the equilibrium constant for this reaction at 25oC?

Zn(s) + Cu+2(aq) -> Cu(s) + Zn+2(aq)

Solution: Use the above equation to determine K. Two electrons are exchanged in the reaction, so n=2

E0 = (0.0257/n)*ln(K)
1.101 V = (0.0257 V/2)*ln(K)
ln(K) = 85.6
K = 1.62*1037
Note that the value is huge: the reaction goes to completion.


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