In an electrochemical cell, the potential difference between two electrodes unde

Question

In an electrochemical cell, the potential difference between two electrodes under standard conditions is known as the standard cell potential (E degree cell). The standard cell potential can be used to identify the overall tendency of a redox reaction to occur spontaneously. The spontaneity of a reaction is identified using the Gibbs free energy G degree G degree is related to E degree cell and are also related to equilibrium constant Keq of the reaction. Select the image to explore the activity that shows how E degree cell, K eq , and G degree are related to each other. In the activity. you should see a triangle, whose three vertices represent Keq, E degree call. and G degree You ca select vertices and determine the relation between them. You can then reset the activity and select the next two quantities. The following values may be useful when solving this tutorial. In the activity, click on the E degree cell and K eq quantities to observe how they are related. Use this relation to calculate K eq 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- rightarrow Cu(s) and Co(s) rightarrow Co2+ (aq) + 2e- The net reaction is Use the given standard reduction potentials in your calculation as appropriate. Express your answer numerically to three significant figures.

Explanation / Answer

Oxidation reaction : anode

Co(s) ----> Co+2 + 2e-

reduction reaction : cathode

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

we know that

EO cell = Eo cathode - Eo anode

so

Eo cell = Eo Cu+2/Cu - EO Co+2/Co

using give values

we get

EO cell = 0.337 + 0.277

Eo cell = 0.614 V

now

we know that

dGo = -nFEo

also

dGo = -RT lnKeq

so

we get

nFEo = RT ln Keq

consider the oxidation reaction

Co(s) ---> Co+2 + 2e-

from the above reaction

we get

n= 2 as two electrons are transferred

so

substituting the values

we get

2 x 96485 x 0.614 = 8.314 x 298 x lnKeq

Keq= 5.87 x 10^20

so the value of equilbrium constant is 5.87 x 10^20

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