The following values may be useful when solving this tutorial. Part A In the act

Question

The following values may be useful when solving this tutorial.

Part A

In the activity, click on the E?cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are

Cu2+(aq)+2e??Cu(s) and Zn(s)?Zn2+(aq)+2e?

The net reaction is

Cu2+(aq)+Zn(s)?Cu(s)+Zn2+(aq)

Use the given standard reduction potentials in your calculation as appropriate.

Constant Value E?Cu 0.337 V E?Zn -0.763V R 8.314 J?mol?1?K?1 F 96,485 C/mol T 298 K

Explanation / Answer

We can always relate the equilibrium constant to cell potentials via Free Energy, dG

the formulas we have:

dG = -n*F*E°cell;

where n = number of electrons being transferred, F = 96500 C/mol, farady consant, E°cell, can be calcualted via E°cell = E°reduction - Eoxidation

Also,

dG = -RT*ln(K);

R = ideal gas constant, 8.314 J/molK, T = temperature, typically 298K, and K the equilibrium constant

so

dG = -RT*ln(K)

dG = -n*F*E°cell

dG1 = dG2

-RT*ln(K) =  -n*F*E°cell

solving for K

K = exp(n*F*E°cell/(RT))

K = exp(n*(96500)*E°cell/(8.314*T))

Now, get E°cell, T in abs, and n, number of e-

n = 2e-,

E°cell = Ereduction - Eoxidation = 0.337 - -0.763 = 1.10 V

K = exp(n*(96500)*E°cell/(8.314*T))

K = exp(2*(96500)*1.10/(8.314*298))

K = 1.637*10^37

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